Part A · the equation of motion · spine

Many oscillators, and the spectral theorem earning its keep

A coupled system is an uncoupled one in the wrong basis — and the right basis exists because two quadratic forms can be diagonalised at once when one of them is definite.

needs 06, 04~2 h slides + 5 h problemsproblems: ./mr problems 07

Unit 06 solved one oscillator to death: damping regimes, resonance, the Green's function. This unit asks what happens when there are $n$ of them and they pull on each other. The answer is not "harder"; it is "the same, after a change of basis", and the whole unit is about why that basis exists, what it is orthogonal with respect to, and how to find it.

The mathematics is a single theorem doing real work. Newton's law for $n$ coupled masses is $M\ddot u = -Ku$ with two symmetric matrices, and the physics asks for a basis diagonalising both at once. That is not always possible for two symmetric matrices. It is possible here, and the reason is that $M$ — the kinetic energy form — is positive definite. Positive definiteness is not a technical footnote; it is the hypothesis the whole subject rests on, and “Three tempting routes, all wrong” shows what breaks without it.

How to read this deck

One system runs through it: $N$ equal masses on $N+1$ equal springs between two walls. Section 1 writes it as a first-order system and finds that machinery is the wrong tool for it; section 2 builds the right one; section 3 solves it exactly and in closed form; section 4 takes the walls away; section 5 bounds its lowest frequency without solving anything. Every number quoted was computed, and each boxed result is checked against dimensions and a case whose answer is already known.

Terms with a dotted underline are glossary entries — click one, or press G for the whole list. N opens these notes, where the longer proofs live under Full proof. R stacks the deck into one readable page.

07 · overview

One chain, five sections

Twenty masses on twenty-one springs, between two walls; $u_j$ is mass $j$'s displacement, so $u \in \mathbb R^{20}$. Pull one mass aside and let go: what you see is a mess. What is happening is twenty independent oscillators, seen in the wrong coordinates.

The running example; masses 4 to 19 are not drawn.

The claim: in the right basis the equations fall apart into twenty copies of unit 06's $\ddot q = -\omega^2 q$, with $\omega_n = 2\sqrt{k/m}\,\sin\big(n\pi/42\big)$ exactly.

how the claim is reached

1. Newton for all masses at once: $M\ddot u = -Ku$, with $M$ the matrix of the kinetic energy $T = \tfrac12\dot u^{\!\top}\!M\dot u$ and $K$ the second derivatives of the potential energy $V$.

2. Every mass swinging at one frequency, $u = v\cos\omega t$, needs $Kv = \omega^2 Mv$.

3. Rescaling by $M^{1/2}$ makes that a symmetric eigenproblem; the spectral theorem gives twenty real $\omega^2$ and a basis of shapes, orthogonal for $T$'s inner product, not the dot product.

4. For the chain, $v_j = \sin j\theta$ fits every mass's equation with $\omega = 2\sqrt{k/m}\sin(\theta/2)$; the far wall needs $\sin 21\theta = 0$, so $\theta = n\pi/21$.

The plan. 1 $\dot y = Ay$ and Jordan · 2 two forms, one basis: steps 1–3 · 3 the chain: step 4, and dispersion · 4 no walls: a symmetry of $V$ is a null vector of $K$ · 5 Rayleigh: any guessed shape bounds the lowest frequency.

Getting around. A dotted underline is a glossary term: click it, or press G for the whole list. A solid underline, like “Twenty masses on twenty-one springs” on this slide, is a piece of the running example: hover it. N opens these notes.

Where each step of the route is done. Step 1 is “Two quadratic forms fall out of the physics”, with “Writing down $M$ and $K$” doing it for the chain. Step 2 is “The generalised eigenvalue problem”. Step 3 is three slides: “The spectral theorem, and its variational proof”, “$M^{1/2}$, and the problem made symmetric” and “What the reduction proves, all at once”, with the orthogonality on “The inner product the modes are orthogonal in”. Step 4 is “Worked: a sine is an eigenvector”. Hover the chain for the full formula with general $N$: $42 = 2(N+1)$ with $N = 20$.

The problem sheet follows the same order: P1–P3 are the linear-systems machinery, P4–P8 the generalised eigenvalue problem and the $M$-inner product, P9–P10 the chain, P11–P12 the zero modes, P13 Rayleigh. Four of them (P2, P7, P8, P13) ask for proofs and several end in a number.

A warning about what this unit is not. It is not about solving $\dot y = Ay$ numerically, and it is not a catalogue of special matrices. It is one structural fact — a pencil of two symmetric forms with one of them definite behaves exactly like a single symmetric matrix — applied until it is second nature. That fact reappears in unit 16 as the inertia tensor's principal axes, in unit 19 as the completeness of the Sturm–Liouville eigenfunctions, and in unit 22 as the principal stresses. Each time, the hypothesis doing the work is definiteness of one form.

section 1

$n$ coordinates, one linear system

Where the two matrices come from, and why the general machinery for $\dot y = Ay$ is the wrong tool for them.

07 · setup

Two quadratic forms fall out of the physics

$n$ masses $m_a$ on a line, displacements $u \in \mathbb R^n$ from an equilibrium of the potential $V$. Unit 04's Taylor theorem at a critical point gives

$$V(u) = V(0) + \tfrac12\,u^{\!\top}\! K u + o(\lVert u\rVert^2),$$

with $K_{ij} = \partial_i\partial_j V(0)$, because $\nabla V(0) = 0$ at an equilibrium. $K$ is symmetric by Clairaut (unit 03) and positive semidefinite when the equilibrium is a minimum.

The force is $F_i = -\partial_i V$ (unit 04), so to first order in $u$

$$M\ddot u = -K u, \qquad M = \mathrm{diag}(m_1,\dots,m_n).$$

definition

The stiffness matrix $K$ is the Hessian of $V$ at the equilibrium. The mass matrix $M$ is the symmetric matrix of the kinetic energy form, $T = \tfrac12\,\dot u^{\!\top}\! M \dot u$.

Both are symmetric; $M$ is positive definite, since $T > 0$ whenever anything is moving.

The small-oscillation approximation is the whole content of "$o(\lVert u\rVert^2)$ dropped": it is exact only in the limit of vanishing amplitude.

In Cartesian displacements $M$ is diagonal. The next slide shows a system where it is not.

07 · setup

A mass matrix that is not diagonal

$M$ is not always diagonal

In Cartesian displacements it is. Change coordinates linearly, $u = Bq$ with $B$ invertible, and $T = \tfrac12\,\dot q^{\!\top}(B^{\!\top}\!MB)\dot q$: any congruent matrix can occur, and every symmetric positive definite matrix is congruent to $I$ (section 2's $M^{1/2}$ shows it). So the theory must handle general $M$.

A double pendulum, in its two angles. Bob 2's velocity is the sum of two pieces, $L_1\dot\theta_1$ and $L_2\dot\theta_2$, each at right angles to its own rod, so at angle $\theta_2-\theta_1$ to each other. Squaring the sum leaves a cross term:

$$\begin{aligned}T = {}&\tfrac12(m_1{+}m_2)L_1^2\dot\theta_1^2 + \tfrac12 m_2L_2^2\dot\theta_2^2\\ &+ m_2L_1L_2\cos(\theta_1{-}\theta_2)\,\dot\theta_1\dot\theta_2 ,\end{aligned}$$

and at small angles $M = \begin{pmatrix}(m_1+m_2)L_1^2 & m_2L_1L_2\\ m_2L_1L_2 & m_2L_2^2\end{pmatrix}$ — symmetric, positive definite, and not diagonal.

Bob 2's velocity in two pieces: $L_1\dot\theta_1$ from rod 1's swing, $L_2\dot\theta_2$ from rod 2's. Their overlap, $L_1\dot\theta_1\,L_2\dot\theta_2\cos(\theta_2-\theta_1)$, is the off-diagonal entry of $M$. Angles exaggerated.

What to look for in the figure. Bob 1 moves on a circle about the pivot, so its velocity is $L_1\dot\theta_1$ at right angles to rod 1. Bob 2 inherits that velocity and adds its own swing about bob 1, $L_2\dot\theta_2$ at right angles to rod 2. The kinetic energy of bob 2 is $\tfrac12m_2$ times the squared length of the sum, and the cross term of that square is the dot product of the two arrows: their lengths times the cosine of the angle between them. That dot product is the off-diagonal entry, and it is there because one coordinate's motion carries the other bob along.

Full derivation: a mass matrix that is genuinely not diagonal

A double pendulum, angles $\theta_1,\theta_2$ from the vertical, rods of lengths $L_1, L_2$ and bobs $m_1, m_2$. The bob positions are $(L_1\sin\theta_1,\,-L_1\cos\theta_1)$ and $(L_1\sin\theta_1 + L_2\sin\theta_2,\,-L_1\cos\theta_1 - L_2\cos\theta_2)$. Differentiating with the chain rule (unit 03) and summing $\tfrac12 m\lVert\dot r\rVert^2$, the cross terms combine into $$T = \tfrac12(m_1+m_2)L_1^2\dot\theta_1^2 + \tfrac12 m_2L_2^2\dot\theta_2^2 + m_2L_1L_2\cos(\theta_1-\theta_2)\,\dot\theta_1\dot\theta_2 .$$ At small angles $\cos(\theta_1-\theta_2) \to 1$, so $M = \begin{pmatrix}(m_1+m_2)L_1^2 & m_2L_1L_2\\ m_2L_1L_2 & m_2L_2^2\end{pmatrix}$, which is symmetric, positive definite (its determinant is $m_1m_2L_1^2L_2^2 > 0$ and its top-left entry is positive) and emphatically not diagonal. The potential $V = (m_1+m_2)gL_1(1-\cos\theta_1) + m_2gL_2(1-\cos\theta_2)$ gives the diagonal $K = \mathrm{diag}\big((m_1+m_2)gL_1,\ m_2gL_2\big)$. So the two forms are diagonal in different bases, which is the whole problem of section 2 in one example.

Unit 12 will show that the same shape, $T = \tfrac12\dot q^{\!\top}M(q)\dot q$, comes out of any change of coordinates, not just linear ones, with $M$ evaluated at the equilibrium. Nothing here waits on that; the linear change above is enough to show a general $M$ must be allowed.

07 · setup

First order, in $2n$ dimensions

Unit 01's move: set $y = (u, \dot u) \in \mathbb R^{2n}$. Then $M\ddot u = -Ku$ becomes

$$\dot y = Ay, \qquad A = \begin{pmatrix} 0 & I_n \\ -M^{-1}K & 0\end{pmatrix},$$

a linear system with constant coefficients. $f(y) = Ay$ is globally Lipschitz with constant $\lVert A\rVert$, so Picard–Lindelöf (unit 01) gives a unique solution for every initial condition, defined for all $t \in \mathbb R$.

theorem · the solution space

The set of solutions of $\dot y = Ay$ on $\mathbb R$ is a vector space of dimension exactly $2n$.

Proof. $y \mapsto y(0)$ is linear from solutions to $\mathbb R^{2n}$; existence makes it onto and uniqueness makes it injective. $\square$

For the chain with $N = 20$ this is a $40 \times 40$ matrix, and the general machinery of the rest of this section would ask for its eigenvalues.

the right tool is not this one

$A$ is not symmetric, and nothing in the general theory of $\dot y = Ay$ knows that it came from two symmetric matrices. Section 2 uses that structure and gets everything the general theory would, in half the dimensions and with real arithmetic.

So why build the general theory at all? Because section 4's zero modes need the degenerate case, and it is the only honest way to see why the degenerate case does not occur for $M$ positive definite and $K$ positive definite.

This is the same reduction unit 01 used to put $m\ddot x = F$ into the hypotheses of Picard–Lindelöf, and unit 02 used to draw phase portraits. Nothing new is being claimed; what is new is that $f$ is now linear, which is what makes an explicit formula for the flow possible.

Why the solution is global. The escape-in-finite-time obstruction of unit 01 needs $\lVert f(y)\rVert$ to grow faster than linearly. Here $\lVert Ay\rVert \le \lVert A\rVert\lVert y\rVert$, so Grönwall (unit 01 again) gives $\lVert y(t)\rVert \le \lVert y(0)\rVert e^{\lVert A\rVert |t|}$: the solution cannot run away in finite time. Linear systems are the model case where the local theorem is global.

Note the dimension count. Each of the $n$ masses needs a position and a velocity, which is $2n$ numbers; and section 2 will produce $n$ modes each contributing a two-dimensional family $q(t) = \alpha\cos\omega t + \beta\sin\omega t$. Two accounts of the same $2n$, which is a check worth keeping in mind.

07 · linear systems

The matrix exponential

definition

The operator norm of $A \in M_n(\mathbb R)$ is $\lVert A\rVert = \sup_{\lVert x\rVert = 1}\lVert Ax\rVert$; it satisfies $\lVert AB\rVert \le \lVert A\rVert\,\lVert B\rVert$.

The matrix exponential is $$e^{A} \;=\; \sum_{k=0}^{\infty}\frac{A^k}{k!},$$ the same series as for a number, with $A^0 = I$.

theorem · it converges

The series converges absolutely for every $A$, and $\lVert e^A\rVert \le e^{\lVert A\rVert}$.

Proof. $\lVert A^k/k!\rVert \le \lVert A\rVert^k/k!$, whose sum is $e^{\lVert A\rVert} < \infty$. $M_n(\mathbb R) \cong \mathbb R^{n^2}$ is complete, and in a complete normed space an absolutely convergent series converges. $\square$

theorem · what it is for

$t \mapsto e^{At}$ is differentiable with $$\frac{\mathrm d}{\mathrm dt}e^{At} = A\,e^{At} = e^{At}A,$$ so $y(t) = e^{At}y_0$ solves $\dot y = Ay$, $y(0) = y_0$ — and by uniqueness it is the only solution.

full proof in the notes (N)

the exponential law needs commuting matrices

With $A = \begin{pmatrix}0&1\\0&0\end{pmatrix}$, $B = \begin{pmatrix}0&0\\1&0\end{pmatrix}$: $e^Ae^B = \begin{pmatrix}2&1\\1&1\end{pmatrix}$, while $e^{A+B} = \begin{pmatrix}\cosh 1&\sinh 1\\ \sinh 1&\cosh 1\end{pmatrix} \approx \begin{pmatrix}1.5431&1.1752\\1.1752&1.5431\end{pmatrix}$.

$e^{A+B} = e^Ae^B$ if $AB = BA$ — which is why $e^{A(t+s)} = e^{At}e^{As}$ is safe.

Full proof that $\frac{\mathrm d}{\mathrm dt}e^{At} = Ae^{At}$

On $|t| \le T$ the $k$-th term has norm at most $\lVert A\rVert^kT^k/k!$, and the termwise $t$-derivative $A^kt^{k-1}/(k-1)!$ has norm at most $\lVert A\rVert^kT^{k-1}/(k-1)!$. Both bounds are summable, so by the Weierstrass M-test both series converge uniformly on $[-T,T]$. A uniformly convergent series of $C^1$ functions whose derivative series also converges uniformly may be differentiated term by term (single-variable analysis, applied entry by entry to the $n^2$ scalar series). Hence $$\frac{\mathrm d}{\mathrm dt}e^{At} = \sum_{k\ge1}\frac{A^kt^{k-1}}{(k-1)!} = A\sum_{j\ge0}\frac{A^jt^j}{j!} = Ae^{At},$$ and pulling $A$ out on the right instead gives $e^{At}A$. Uniqueness of the solution is unit 01's Picard–Lindelöf with Lipschitz constant $\lVert A\rVert$.

Full proof that $AB = BA \Rightarrow e^{A+B} = e^Ae^B$

Both series converge absolutely, so their Cauchy product may be rearranged: $e^Ae^B = \sum_{k}\sum_{i+j=k}\frac{A^iB^j}{i!\,j!}$. When $A$ and $B$ commute the binomial theorem holds for them, $(A+B)^k = \sum_{i+j=k}\binom{k}{i}A^iB^j$, and dividing by $k!$ gives exactly the inner sum. Without commutativity the binomial theorem is false at $k = 2$ already: $(A+B)^2 = A^2 + AB + BA + B^2 \ne A^2 + 2AB + B^2$.

The counterexample, checked. $A^2 = 0$ so $e^A = I + A = \begin{pmatrix}1&1\\0&1\end{pmatrix}$, and likewise $e^B = \begin{pmatrix}1&0\\1&1\end{pmatrix}$; the product is $\begin{pmatrix}2&1\\1&1\end{pmatrix}$. For $A + B = \begin{pmatrix}0&1\\1&0\end{pmatrix}$, the square is $I$, so the even terms sum to $\cosh 1\,I$ and the odd ones to $\sinh 1\,(A+B)$. The traces differ, $3$ against $2\cosh 1 = 3.0862$, so no amount of rearranging can reconcile them.

Historically this failure is the beginning of Lie theory: the correct statement is the Baker–Campbell–Hausdorff formula, whose first correction term is $\tfrac12[A,B]$. This course does not need it, but unit 17 meets the commutator again as the Poisson bracket, and unit 33 as the curvature.

07 · linear systems

When $A$ is diagonalisable

theorem

If $A = P\Lambda P^{-1}$ with $\Lambda = \mathrm{diag}(\mu_1,\dots,\mu_n)$, then $$e^{At} = P\,\mathrm{diag}\big(e^{\mu_1 t},\dots,e^{\mu_n t}\big)\,P^{-1},$$ and every solution of $\dot y = Ay$ is $y(t) = \sum_i c_i e^{\mu_i t}w_i$ with $w_i$ the columns of $P$.

Proof. $A^k = P\Lambda^kP^{-1}$, and $P(\cdot)P^{-1}$ is continuous and linear, so it passes through the sum. $\square$

Complex $\mu$ is not a problem: unit 06's complex exponential is used exactly as it was there, and real initial data gives real solutions because the complex eigenvalues come in conjugate pairs.

one oscillator, as a check

$\ddot x = -\omega_0^2x$ has $A = \begin{pmatrix}0&1\\-\omega_0^2&0\end{pmatrix}$, eigenvalues $\pm i\omega_0$, eigenvectors $(1, \pm i\omega_0)$. Then

$$e^{At} = \begin{pmatrix}\cos\omega_0 t & \dfrac{\sin\omega_0 t}{\omega_0}\\[4pt] -\omega_0\sin\omega_0 t & \cos\omega_0 t\end{pmatrix}.$$

First column: released from rest at $x = 1$. Second column: struck from $x=0$ with unit speed — unit 06's Green's function $G(t) = \sin\omega_0t/\omega_0$, recovered.

Dimensions: $[A] = \mathrm s^{-1}$, so $At$ is dimensionless and the series makes sense. The entry $\sin\omega_0t/\omega_0$ carries $\mathrm s$, as a map from velocity to position must.

The example in full. $\det(A - \mu I) = \mu^2 + \omega_0^2$, so $\mu = \pm i\omega_0$; $(A - i\omega_0)w = 0$ gives $w = (1, i\omega_0)$. With $P = \begin{pmatrix}1&1\\ i\omega_0&-i\omega_0\end{pmatrix}$ and $P^{-1} = \tfrac12\begin{pmatrix}1&1/(i\omega_0)\\ 1&-1/(i\omega_0)\end{pmatrix}$, the product $P\,\mathrm{diag}(e^{i\omega_0t},e^{-i\omega_0t})\,P^{-1}$ has $(1,1)$ entry $\tfrac12(e^{i\omega_0t}+e^{-i\omega_0t}) = \cos\omega_0t$ and $(1,2)$ entry $\tfrac{1}{2i\omega_0}(e^{i\omega_0t}-e^{-i\omega_0t}) = \sin\omega_0t/\omega_0$. The bottom row is the $t$-derivative of the top, as it must be, since the second component of $y$ is $\dot x$.

Limit check. As $\omega_0 \to 0$ the matrix tends to $\begin{pmatrix}1&t\\0&1\end{pmatrix}$, which is free motion $x = x_0 + v_0t$ — and it is also the degenerate case of the next slide, since at $\omega_0 = 0$ the two eigenvalues collide at $0$. The free particle is the simplest Jordan block in mechanics, and section 4 meets it again as a zero mode.

A caution about $P$. Diagonalising is a statement about a basis of eigenvectors, and $P$ is the matrix whose columns are that basis. Nothing forces $P$ to be orthogonal; it is orthogonal exactly when $A$ is symmetric, which is the spectral theorem, and which is why section 2 works so hard to get a symmetric matrix.

07 · linear systems

The degenerate case: Jordan, and $t\,e^{\lambda t}$

Cited. Every $A \in M_n(\mathbb C)$ is similar to a Jordan form: blocks $\lambda I + N$ with $N$ the nilpotent shift ($N^p = 0$ for some $p$; this $N$ is a matrix, not the chain's number of masses). Proved in any linear algebra text; the only case we use is verified below.

theorem · exponential of a block

If $A = \lambda I + N$ with $N^p = 0$, then $$e^{At} = e^{\lambda t}\Big(I + Nt + \dots + \tfrac{N^{p-1}t^{p-1}}{(p-1)!}\Big),$$ a polynomial in $t$ times $e^{\lambda t}$.

Proof. $\lambda I$ commutes with $N$, so $e^{At} = e^{\lambda t}e^{Nt}$, and that series stops at $k = p-1$. $\square$

So a repeated eigenvalue gives $te^{\lambda t}$, $t^2e^{\lambda t}$, … — but only a defective one does.

worked · critical damping, again

Unit 06's $\ddot x + \gamma\dot x + \omega_0^2x = 0$ at $\gamma = 2\omega_0$: $A = \begin{pmatrix}0&1\\-\omega_0^2&-2\omega_0\end{pmatrix}$, with $\det(A-\mu I) = (\mu+\omega_0)^2$.

$N = A + \omega_0I = \begin{pmatrix}\omega_0&1\\-\omega_0^2&-\omega_0\end{pmatrix}$ has $N^2 = 0$ (each entry of $N^2$ is a difference of two equal products). Hence $$e^{At} = e^{-\omega_0t}\begin{pmatrix}1+\omega_0t&t\\ -\omega_0^2t&1-\omega_0t\end{pmatrix},$$ and the $te^{-\omega_0t}$ of unit 06 is the nilpotent part, not a lucky guess.

Plotted on the next slide.

Why $A$ is defective here. $(A+\omega_0I)w = 0$ reads $\omega_0w_1 + w_2 = 0$, a single equation, so the eigenspace is one-dimensional while the eigenvalue has algebraic multiplicity two. That is precisely the definition of defective, and it is exactly the circumstance in which unit 06's two-parameter solution family could not be written as $c_1e^{\mu_1t} + c_2e^{\mu_2t}$.

The dimension check. $[\omega_0] = \mathrm s^{-1}$, so the matrix entries have dimensions $\begin{pmatrix}1 & \mathrm s\\ \mathrm s^{-2} & 1\end{pmatrix}$ — position from position, position from velocity, velocity from position, velocity from velocity. Every propagator in this unit has that shape.

Where this will and will not bite. Section 2 proves that the undamped system $M\ddot u = -Ku$ with $M$ positive definite never produces a defective problem in $u$-space: repeated $\omega^2$ still has a full eigenspace. Defectiveness returns in exactly two places: damping (here), and zero modes (section 4), where $\mu = 0$ is a genuine $2\times2$ Jordan block in the $2n$-dimensional first-order system, giving $q(t) = \alpha + \beta t$. Everything else oscillates.

A remark for later: defective eigenvalues are fragile. Perturb the nilpotent $\begin{pmatrix}0&1\\0&0\end{pmatrix}$ to $\begin{pmatrix}0&1\\\varepsilon&0\end{pmatrix}$ and the characteristic polynomial becomes $\mu^2 - \varepsilon$, so the double eigenvalue $0$ splits into $\pm\sqrt\varepsilon$ — a shift far larger than the size $\varepsilon$ of the perturbation. For a symmetric matrix and a symmetric perturbation of operator norm $\varepsilon$, every eigenvalue moves by at most $\varepsilon$ (a consequence of section 5's min–max characterisation, since the quotient $u^{\!\top}Su/u^{\!\top}u$ itself moves by at most $\varepsilon$), which is one more thing the spectral theorem buys.

07 · linear systems · example

Critical damping, plotted

The two columns of $e^{At}$ on the last slide are two motions; the first row of each is $x(t)$:

released from rest at $x = 1$: $\;x = (1+\omega_0t)\,e^{-\omega_0t}$;
struck at $x = 0$ with unit speed: $\;x = t\,e^{-\omega_0t}$.

Each is $e^{-\omega_0t}$ times a degree-one polynomial, and the degree-one part is the block's $Nt$.

checks

At $t = 0$ the released curve has slope $0$ and the struck one slope $1$, as the initial data demand. The struck curve peaks where $\tfrac{\mathrm d}{\mathrm dt}(te^{-\omega_0t}) = (1-\omega_0t)e^{-\omega_0t} = 0$: at $\omega_0t = 1$, height $1/(e\,\omega_0)$.

The closed form matches a direct summation of the series at $\omega_0 = 2$, $t = 0.7$ to below $10^{-15}$.

Both columns in units of $\omega_0$ (the struck curve as $\omega_0x$, per unit speed). The dashed $e^{-\omega_0t}$ is what the eigenvalue alone would give: the polynomial factor holds the released body up and pushes the struck one out.

section 2

Two quadratic forms, one basis

The generalised eigenvalue problem, and the single hypothesis — $M$ positive definite — that makes it behave.

07 · the pencil

The generalised eigenvalue problem

Look for a motion in which every coordinate oscillates with the same frequency and a fixed shape: $u(t) = v\,f(t)$ with $v \in \mathbb R^n$ constant. Then $M\ddot u = -Ku$ reads $\ddot f\,Mv = -f\,Kv$, which forces $\ddot f = -\lambda f$ and

definition

The generalised eigenvalue problem for the pair $(K, M)$ is $$Kv = \lambda\,Mv, \qquad v \ne 0 .$$ The family $K - \lambda M$ is the matrix pencil; its characteristic polynomial is $p(\lambda) = \det(K - \lambda M)$.

A solution $(\lambda, v)$ with $\lambda = \omega^2 \ge 0$ is a normal mode: shape $v$, normal frequency $\omega$, motion $u(t) = v\,(\alpha\cos\omega t + \beta\sin\omega t)$.

degree, and therefore a count

$\det(K-\lambda M) = \det(M)\det(M^{-1}K - \lambda I)$ and $\det M \ne 0$, so $p$ has degree exactly $n$: $n$ roots in $\mathbb C$, with multiplicity (fundamental theorem of algebra, cited; it follows from Liouville's theorem in unit 20).

That is a count, not yet a theorem about physics. Three things still have to be proved, and none of them follows from $K$ and $M$ merely being symmetric:

every root is real;
every root is $\ge 0$ when $K$ is positive semidefinite, so $\omega = \sqrt\lambda$ is real;
the eigenvectors form a basis — no defective case, so $n$ coordinates really do give $n$ modes.

On the chain, $M = mI$ and $\lambda = \omega^2$ is an ordinary eigenvalue of $K/m$. The general case is the one to get right.

Why the ansatz is not a loss of generality. It looks like a guess, and at this point it is. What justifies it is the theorem on “What the reduction proves, all at once”: the modes form a basis of $\mathbb R^n$, so every solution is a superposition of them. The ansatz finds $n$ two-parameter families, $2n$ constants in all, which matches the dimension of the solution space counted on “First order, in $2n$ dimensions”. Nothing is missed.

Why $\ddot f = -\lambda f$ is forced. From $\ddot f\,Mv = -f\,Kv$ with $v$ fixed and $Mv \ne 0$ (as $M$ is invertible and $v \ne 0$): at any $t$ with $f(t) \ne 0$, the vectors $Mv$ and $Kv$ are parallel, so $Kv = \lambda Mv$ for some scalar $\lambda$ independent of $t$, and then $\ddot f = -\lambda f$.

Terminology. "Pencil" is nineteenth-century projective-geometry language for a one-parameter linear family — a pencil of lines through a point, a pencil of conics. The pair $(K, M)$ is called a symmetric definite pencil when $M$ is positive definite, and that is the hypothesis every good theorem in this unit carries.

07 · the pencil

Three tempting routes, all wrong

1 · "$M^{-1}K$ is symmetric"

It is not. For the two-mass example, $$M^{-1}K = \begin{pmatrix}2&-1\\-\tfrac14&\tfrac12\end{pmatrix}.$$ So the spectral theorem cannot be applied to it directly. Its eigenvalues are real, but that needs proof.

2 · "diagonalise both with one orthogonal $Q$"

Possible only if $MK = KM$. Here $MK = \begin{pmatrix}2&-1\\-4&8\end{pmatrix}$ and $KM = \begin{pmatrix}2&-4\\-1&8\end{pmatrix}$: different. Orthogonal simultaneous diagonalisation is false, yet the physics still works — so the right notion is not similarity.

3 · "any two symmetric forms will do"

Take $M = \begin{pmatrix}1&0\\0&-1\end{pmatrix}$, $K = \begin{pmatrix}0&1\\1&0\end{pmatrix}$, both symmetric, neither definite. Then $$\det(K-\lambda M) = -\lambda^2 - 1,$$ whose roots are $\pm i$. No real $\lambda$ at all, so no real basis diagonalises both.

The lesson of the third box: definiteness of one form is the hypothesis. The right notion for the other two is congruence, $C^{\!\top}\!(\cdot)C$, not similarity $P^{-1}(\cdot)P$ — a change of basis for a bilinear form, not for a linear map.

Box 2, in full. The standard theorem is: real symmetric $S, T$ are simultaneously diagonalisable by a single orthogonal $Q$ if and only if $ST = TS$. ($\Leftarrow$ is the usual argument on joint eigenspaces; $\Rightarrow$ is immediate since diagonal matrices commute.) Our $M$ and $K$ do not commute, so that theorem is unavailable, and yet the rest of this section produces a basis diagonalising both. The difference is that the basis is orthonormal in a different inner product — which is “The inner product the modes are orthogonal in”.

Box 3, in full. Suppose some invertible $C$ gave $C^{\!\top}MC = D_1$ and $C^{\!\top}KC = D_2$ both diagonal. Then $\det(K - \lambda M) = \det(C^{-\top})\det(D_2 - \lambda D_1)\det(C^{-1})$, and $\det(D_2-\lambda D_1) = \prod_i\big((D_2)_{ii} - \lambda (D_1)_{ii}\big)$. Every $(D_1)_{ii} \ne 0$, because $\det D_1 = (\det C)^2\det M \ne 0$, so each factor has the real root $(D_2)_{ii}/(D_1)_{ii}$. Our pencil's roots are $\pm i$, so no such $C$ exists. The physical reading: a "kinetic energy" that can be negative admits motions that are neither oscillations nor exponentials but a genuinely complex mixture. Such pencils occur in real physics — in plasma stability and in the Hamiltonian normal-form theory behind unit 17 — and they are hard precisely because this unit's theorem is unavailable.

Why the distinction similarity/congruence matters. A matrix can represent either a linear map or a bilinear form, and the two transform differently under a change of basis: $P^{-1}AP$ for a map, $C^{\!\top}\!BC$ for a form. $M$ and $K$ are forms — energies — so congruence is their natural equivalence. Unit 16 makes the distinction permanent by calling one a $(1,1)$-tensor and the other a $(0,2)$-tensor, and unit 26 will note that $C^{\!\top}\eta\,C = \eta$ is the definition of the Lorentz group.

07 · the spectral theorem

The spectral theorem, and its variational proof

spectral theorem

Let $S \in M_n(\mathbb R)$ be symmetric. Then $\mathbb R^n$ has an orthonormal basis $w_1,\dots,w_n$ of eigenvectors of $S$, with real eigenvalues $\mu_1 \le \dots \le \mu_n$. Equivalently $S = Q\,\mathrm{diag}(\mu)\,Q^{\!\top}$ with $Q$ orthogonal.

proof, in outline

The function $u \mapsto u^{\!\top}\!Su$ is continuous on the unit sphere, which is compact, so it attains a maximum at some $w$. Lagrange multipliers (unit 04) for $\max u^{\!\top}\!Su$ subject to $u^{\!\top}u = 1$ give $Sw = \mu w$. Then $S$ maps $w^{\perp}$ into itself, because $\langle Sx, w\rangle = \langle x, Sw\rangle = \mu\langle x, w\rangle = 0$, and the restriction is symmetric. Induct on the dimension. $\square$

every step written out in the notes (N)

The proof in two dimensions, for $S = \begin{pmatrix}2&-\frac12\\-\frac12&\frac12\end{pmatrix}$ (it returns on the $M^{1/2}$ slide). Where $u^{\!\top}\!Su$ is largest or smallest on the circle, its level curve touches the circle, so its gradient $2Su$ points along the radius $u$: that is $Su = \mu u$.

What to look for in the figure. The level curves of $u^{\!\top}\!Su$ are ellipses centred at the origin. Follow $u$ round the unit circle: $u^{\!\top}\!Su$ rises and falls, and at its maximum the level curve through $u$ cannot cross the circle (it would then reach higher values on it), so it touches it. Touching curves share a normal; the normal to the circle is $u$, the normal to the level curve is $\nabla(u^{\!\top}\!Su) = 2Su$, hence $Su = \mu u$ — that is Lagrange's condition, drawn. The same happens at the minimum, at right angles to the maximum: the second eigenvector, which the induction step finds by repeating the argument on $w^\perp$, a line here. The dashed curve $u^{\!\top}\!Su = 1$ crosses the circle at four points, none of them stationary.

Full proof

Induct on $n$. For $n = 1$ there is nothing to prove. For $n > 1$, let $g(u) = u^{\!\top}\!Su$ and $h(u) = u^{\!\top}u - 1$. The sphere $\{h = 0\}$ is closed and bounded, hence compact (Heine–Borel), and $g$ is continuous, so $g$ attains a maximum at some $w$ with $\lVert w\rVert = 1$. Since $\nabla h(w) = 2w \ne 0$, the constraint qualification of unit 04's Lagrange multiplier theorem holds, so there is $\mu$ with $\nabla g(w) = \mu\nabla h(w)$. Now $\nabla g(u) = 2Su$ for symmetric $S$ — this is where symmetry enters, and only here — so $2Sw = 2\mu w$, i.e. $Sw = \mu w$ with $\mu = w^{\!\top}\!Sw \in \mathbb R$.

Let $W = w^\perp$, of dimension $n-1$. For $x \in W$, $\langle Sx, w\rangle = \langle x, Sw\rangle = \mu\langle x,w\rangle = 0$, so $Sx \in W$: $S$ restricts to an operator on $W$, and it is symmetric there because the restriction of a symmetric bilinear form to a subspace is symmetric. Choose an orthonormal basis of $W$; in it the restriction is a symmetric $(n-1)\times(n-1)$ matrix, and the inductive hypothesis gives an orthonormal eigenbasis of $W$. Adjoining $w$ gives one of $\mathbb R^n$. $\square$

Why $\nabla(u^{\!\top}\!Su) = (S + S^{\!\top})u$ in general. Expand $(u+h)^{\!\top}S(u+h) = u^{\!\top}Su + u^{\!\top}Sh + h^{\!\top}Su + h^{\!\top}Sh$. The middle terms are scalars, so $u^{\!\top}Sh = h^{\!\top}S^{\!\top}u$, and the linear part in $h$ is $h^{\!\top}(S + S^{\!\top})u$; the quadratic part is $O(\lVert h\rVert^2)$. So by unit 03's definition of the total derivative, $\nabla g(u) = (S+S^{\!\top})u$, which is $2Su$ when $S$ is symmetric. The antisymmetric part of $S$ contributes nothing to the form, which is why only the symmetric part was ever relevant.

07 · the spectral theorem

Two things the proof leaves behind

Two things to notice about that proof, both used later:

the eigenvector it produces is a maximiser of a ratio — on the unit sphere $u^{\!\top}\!Su$ is the ratio $u^{\!\top}\!Su/u^{\!\top}u$, which is section 5's Rayleigh quotient, already here;
it never leaves $\mathbb R$, so reality of the eigenvalues is a conclusion, not an assumption.

the physics of the statement

A symmetric matrix is a quadratic form $u \mapsto u^{\!\top}\!Su$, that is, an energy. The theorem says any energy is, in suitable orthonormal coordinates, a sum $\sum\mu_i x_i^2$ of independent one-dimensional energies. Every "principal axes" statement in physics — of inertia, of stress, of strain, of a conic — is this one sentence.

In the last slide's figure the touching points sit on the ellipses' axes: those axes are the principal axes, and $w_1$, $w_2$ point along them.

07 · the reduction

$M^{1/2}$, and the problem made symmetric

definition and lemma

For $M$ symmetric positive definite, write $M = Q\,\mathrm{diag}(\mu_i)\,Q^{\!\top}$ with all $\mu_i > 0$ and set $$M^{1/2} = Q\,\mathrm{diag}(\sqrt{\mu_i})\,Q^{\!\top}.$$ This matrix square root is symmetric positive definite, satisfies $(M^{1/2})^2 = M$, and is the only positive definite matrix that does.

uniqueness proved in the notes (N)

theorem · the reduction

Put $S = M^{-1/2}KM^{-1/2}$, which is symmetric. Then $$Kv = \lambda Mv \iff Sw = \lambda w, \qquad w = M^{1/2}v ,$$ and $v \mapsto M^{1/2}v$ is a bijection of eigenvectors.

Proof. Multiply $Kv = \lambda Mv$ on the left by $M^{-1/2}$ and insert $I = M^{-1/2}M^{1/2}$ in the middle: $M^{-1/2}KM^{-1/2}\,(M^{1/2}v) = \lambda\,(M^{1/2}v)$. Each step is reversible, since $M^{1/2}$ is invertible. Symmetry: $S^{\!\top} = M^{-1/2}K^{\!\top}M^{-1/2} = S$. $\square$

what has been bought

The pencil problem is now an ordinary symmetric eigenvalue problem, so the spectral theorem applies to it whole.

On the two-mass example: $M^{1/2} = \mathrm{diag}(1,2)$, so $$S = \begin{pmatrix}1&0\\0&\tfrac12\end{pmatrix}\!\begin{pmatrix}2&-1\\-1&2\end{pmatrix}\!\begin{pmatrix}1&0\\0&\tfrac12\end{pmatrix} = \begin{pmatrix}2&-\tfrac12\\-\tfrac12&\tfrac12\end{pmatrix},$$ symmetric, where $M^{-1}K$ was not.

Physically, $w = M^{1/2}v$ is the displacement measured in units of $1/\sqrt{\text{mass}}$ — "mass-weighted coordinates", in which the kinetic energy is $\tfrac12\lVert\dot w\rVert^2$ with no matrix left in it.

Full proof of uniqueness of the positive definite square root

Suppose $R$ is positive definite with $R^2 = M$. Let $\mu > 0$ be an eigenvalue of $M$ and $E_\mu$ its eigenspace. If $Rx = \rho x$ then $Mx = R^2x = \rho^2x$, so every eigenvector of $R$ is an eigenvector of $M$ with eigenvalue $\rho^2$, and $\rho > 0$ since $R$ is positive definite. By the spectral theorem $R$ has an eigenbasis, so $\mathbb R^n$ decomposes into $R$-eigenspaces, each contained in some $M$-eigenspace, and $\rho = \sqrt\mu$ is determined by which. Hence $R$ acts as $\sqrt\mu$ on all of $E_\mu$ — the $R$-eigenspaces inside $E_\mu$ all carry the same $\rho$ and therefore fill it — which is exactly how $M^{1/2}$ acts. So $R = M^{1/2}$. $\square$

What could go wrong without positive definiteness. $I$ has infinitely many symmetric square roots: every reflection $\mathrm{diag}(\pm1,\dots,\pm1)$ conjugated by an orthogonal matrix. Insisting the root be positive definite picks out one. And a matrix with a negative eigenvalue has no real symmetric square root at all (the square of a symmetric matrix is positive semidefinite), which is why $M$ positive definite is not a convenience here but the point.

An alternative route avoids square roots: factor $M = C^{\!\top}C$ by Cholesky and set $S = C^{-\top}KC^{-1}$, again symmetric. It is the route numerical work takes, because a Cholesky factorisation costs a small fraction of an eigendecomposition. The theory is identical; the conclusions below hold for any $C$ with $M = C^{\!\top}C$, and $C = M^{1/2}$ is just the symmetric choice. What is not symmetric is $M^{-1}K = C^{-1}(C^{-\top}K)$: both inverse factors sit on the left, where the symmetric version splits them, $C^{-\top}KC^{-1}$, one on each side. The two are similar, $M^{-1}K = C^{-1}\,(C^{-\top}KC^{-1})\,C$, which is why $M^{-1}K$ has the same real eigenvalues — the proof that box 1 of “Three tempting routes, all wrong” asked for.

Dimensions. $[M^{1/2}] = \mathrm{kg}^{1/2}$ and $[K] = \mathrm{kg\,s^{-2}}$, so $[S] = \mathrm{s^{-2}}$: an eigenvalue of $S$ is a squared frequency, as required. In the two-mass example the entries of $S$ are in $\mathrm s^{-2}$ with $k = 1\ \mathrm{N\,m^{-1}}$ and masses in kg.

07 · the reduction

What the reduction proves, all at once

theorem · normal modes exist, as many as there are coordinates

Let $M$ be symmetric positive definite and $K$ symmetric, both $n\times n$. Then:

all $n$ roots of $\det(K - \lambda M)$ are real;
there is a basis $v_1,\dots,v_n$ of $\mathbb R^n$ with $Kv_i = \lambda_iMv_i$ — no defective case;
if moreover $K$ is positive semidefinite then every $\lambda_i \ge 0$, so $\omega_i = \sqrt{\lambda_i}$ is real; and $\lambda_i > 0$ for all $i$ exactly when $K$ is positive definite.

Proof. By the reduction, the roots and eigenvectors are those of the symmetric $S$, transported by the invertible $M^{1/2}$. (1) and (2) are the spectral theorem for $S$. For (3), $\lambda = w^{\!\top}Sw/\lVert w\rVert^2 = v^{\!\top}Kv/(v^{\!\top}Mv)$ with $v = M^{-1/2}w$, and both quadratic forms on the right are $\ge 0$. $\square$

dimensions

$[K]/[M] = (\mathrm{N\,m^{-1}})/\mathrm{kg} = \mathrm{s^{-2}}$. So $\lambda$ is a squared frequency and $\omega$ a frequency, for any $M$ and $K$ whatever.

limit: one mass

$M = (m)$, $K = (k)$: one root $\lambda = k/m$, one mode $v = (1)$, $\omega = \sqrt{k/m}$. Unit 06's oscillator, recovered.

no complex frequencies

$\omega^2$ real always. $\omega^2 < 0$ is possible — an unstable equilibrium, $e^{\pm|\omega|t}$, unit 02's saddle — but an $\omega$ with a nonzero real and imaginary part never occurs.

Read item 2 twice. It is the part that does not follow from counting. A general matrix with a repeated eigenvalue may be defective, and then the solutions contain $te^{\lambda t}$ — growing amplitudes with no energy input, which would be physical nonsense for a conservative system. The theorem says that cannot happen: repeated $\omega^2$ still has an eigenspace of the full multiplicity, and a degenerate pair of modes is simply a two-dimensional space of shapes all oscillating at the same frequency. The chain in section 3 has no degeneracies; a square drum-frame or a symmetric molecule has plenty, and they are the fingerprint of a symmetry group.

Why "no complex frequencies" is the right slogan. Write a solution as $e^{i\omega t}$. If $\omega = \alpha + i\beta$ with both parts nonzero, the motion is an oscillation with an exponentially changing amplitude. That is what a damped or a driven-unstable system does, and unit 06 met it. The present theorem forbids it for $M\ddot u = -Ku$ with $M,K$ symmetric and $M$ positive definite: the only options are genuine oscillation ($\lambda > 0$), pure exponential growth and decay ($\lambda < 0$), and linear drift ($\lambda = 0$). Energy conservation is the physical reason, and symmetry of $K$ is its algebraic shadow — a non-symmetric $K$ is exactly a force field that is not a gradient, which unit 04 showed can do net work around a loop.

The stable case. $K$ positive definite $\iff$ the equilibrium is a strict local minimum of $V$ to second order $\iff$ all $\omega_i > 0$ $\iff$ every small disturbance of the linearised system stays small. The nonlinear version — the Lagrange–Dirichlet theorem: a strict local minimum of $V$ is a stable equilibrium — is already within reach. Full proof: $E = T + V - V(u^*)$ is conserved, vanishes at the equilibrium and is positive on a punctured neighbourhood of it in phase space (as $T > 0$ unless $\dot u = 0$, and $V > V(u^*)$ unless $u = u^*$), so it is a Lyapunov function and unit 02's direct method gives stability. Note it needs a strict minimum, which is weaker than $K$ positive definite: $V = u^4$ qualifies.

07 · the $M$-inner product

The inner product the modes are orthogonal in

definition

For $M$ symmetric positive definite, the $M$-inner product on $\mathbb R^n$ is $$\langle u, v\rangle_M = u^{\!\top}\!Mv .$$ It is bilinear, symmetric and positive definite — the three axioms of an inner product — so it is one, and $\lVert u\rVert_M = \sqrt{u^{\!\top}Mu}$ is a norm.

The physics: $\tfrac12\langle\dot u,\dot u\rangle_M$ is the kinetic energy. The modes are orthogonal in the geometry that the energy defines, not the one the coordinate labels happen to define.

theorem · $M$-orthogonality

If $Kv_i = \lambda_iMv_i$, $Kv_j = \lambda_jMv_j$ and $\lambda_i \ne \lambda_j$, then $\langle v_i,v_j\rangle_M = 0$ and $v_i^{\!\top}\!Kv_j = 0$.

Proof. $\lambda_i\,v_j^{\!\top}Mv_i = v_j^{\!\top}Kv_i = (Kv_j)^{\!\top}v_i = \lambda_j\,v_j^{\!\top}Mv_i$, using $K = K^{\!\top}$ and $M = M^{\!\top}$. So $(\lambda_i-\lambda_j)v_j^{\!\top}Mv_i = 0$. $\square$

Repeated $\lambda$: the eigenspace is a genuine subspace, so Gram–Schmidt in $\langle\cdot,\cdot\rangle_M$ makes it $M$-orthonormal. Hence a full $M$-orthonormal basis of modes, $\langle v_i,v_j\rangle_M = \delta_{ij}$.

Equivalently: $M^{1/2}$ carries $\langle\cdot,\cdot\rangle_M$ to the Euclidean product, and the $w_i$ are Euclidean-orthonormal by the spectral theorem.

Why an inner product and not just "a useful bilinear form". Because every consequence of the inner-product axioms is now available for free: Cauchy–Schwarz, the Pythagoras identity, Gram–Schmidt, orthogonal projection, Bessel's inequality, and the fact that an orthonormal set is linearly independent. Section 5's Rayleigh argument and unit 19's Fourier expansions both use those, and they use them in $\langle\cdot,\cdot\rangle_M$ without a word of extra proof.

Positive definiteness is again the hypothesis. If $M$ had a negative eigenvalue, $\langle\cdot,\cdot\rangle_M$ would be an indefinite form, not an inner product; orthogonality would no longer imply independence, and a nonzero vector could be "orthogonal to itself". Unit 26 takes exactly that step deliberately: the Minkowski form is symmetric and indefinite, self-orthogonal vectors are the light rays, and a great deal of the strangeness of relativity is the loss of the axiom dropped here.

A consistency check on the proof. It uses only symmetry of $K$ and $M$, not definiteness — so $M$-orthogonality of modes with distinct eigenvalues is true even for an indefinite $M$. What fails without definiteness is everything else: reality of $\lambda$, the existence of a basis, and the very claim that $\langle\cdot,\cdot\rangle_M$ is an inner product.

07 · the $M$-inner product · example

Worked: two masses, and the orthogonality that is not Euclidean

worked · the two-mass example

1. The matrices. $V = \tfrac12 u_1^2 + \tfrac12(u_2-u_1)^2 + \tfrac12u_2^2$, so $K = \begin{pmatrix}2&-1\\-1&2\end{pmatrix}$, $M = \begin{pmatrix}1&0\\0&4\end{pmatrix}$ (SI units).

2. The characteristic polynomial. $\det(K-\lambda M) = (2-\lambda)(2-4\lambda) - 1 = 4\lambda^2 - 10\lambda + 3$, so $$\lambda = \tfrac{10\pm\sqrt{52}}{8} = \tfrac{5\pm\sqrt{13}}{4} = 0.348612,\ 2.151388\ \ \mathrm{s^{-2}} .$$

3. The shapes. Row 1 of $(K-\lambda M)v = 0$ reads $(2-\lambda)v_1 = v_2$, so $$v^{(1)} = \big(1,\ \tfrac{3+\sqrt{13}}{4}\big), \qquad v^{(2)} = \big(1,\ \tfrac{3-\sqrt{13}}{4}\big),$$ numerically $(1,\ 1.651388)$ and $(1,\ -0.151388)$: together, then opposed.

4. The two inner products. $v^{(1)}\!\cdot v^{(2)} = 1 + \tfrac{(3+\sqrt{13})(3-\sqrt{13})}{16} = \tfrac34 \ne 0$, but $\langle v^{(1)},v^{(2)}\rangle_M = 1 + 4\big(-\tfrac14\big) = 0 .$

The two shapes, as displacement arrows (to scale within each row). Low mode: together, the heavy mass further. High mode: opposed, the heavy mass nearly still.

$\omega_1 = 0.590434$, $\omega_2 = 1.466761\ \mathrm{s^{-1}}$; periods $10.642$ s and $4.2837$ s.

Every number above, independently. The exact roots are $(5\pm\sqrt{13})/4$; numerically $0.3486121811$ and $2.1513878189$, and a direct eigendecomposition of $S = M^{-1/2}KM^{-1/2} = \begin{pmatrix}2&-\frac12\\-\frac12&\frac12\end{pmatrix}$ returns the same two to fourteen digits. $\omega_1 = 0.5904339$, $\omega_2 = 1.4667610\ \mathrm{s^{-1}}$; $T_1 = 10.64164$ s, $T_2 = 4.283714$ s, ratio $\omega_2/\omega_1 = 2.4842087$. The Euclidean norms are $\lVert v^{(1)}\rVert = \sqrt{1+1.651388^2} = 1.930565$ and $\lVert v^{(2)}\rVert = \sqrt{1+0.151388^2} = 1.011394$, giving $\cos\theta = 0.75/1.952562 = 0.384111$ and $\theta = 67.41^\circ$.

Why the $M$-product came out exactly zero, exactly. With $v^{(i)} = (1, 2-\lambda_i)$, $\langle v^{(1)},v^{(2)}\rangle_M = 1 + 4(2-\lambda_1)(2-\lambda_2)$, and $(2-\lambda_1)(2-\lambda_2) = 4 - 2(\lambda_1+\lambda_2) + \lambda_1\lambda_2 = 4 - 5 + \tfrac34 = -\tfrac14$ by Vieta. So the product is $1 - 1 = 0$ identically, not to rounding — which is the theorem, seen in the arithmetic.

Physical reading of the shapes. In the lowest mode the two masses move together, and it is the heavy one that swings further, by a factor $1.651$: the middle spring is then the least stretched of the three, which is what keeps the frequency low. Check it on mass 1: the restoring force is $-2u_1 + u_2 = -2 + 1.651 = -0.349$, and $\lambda_1u_1 = 0.349$. ✓ On mass 2: $u_1 - 2u_2 = 1 - 3.303 = -2.303$, and $\lambda_1m_2u_2 = 0.349\times4\times1.651 = 2.303$. ✓ In the upper mode they move oppositely and the light mass carries almost all of the motion: $m_2$ is four times as heavy and moves only $0.151$ as far. The two momentum amplitudes, $m_1v^{(2)}_1 = 1$ against $m_2v^{(2)}_2 = -0.606$ (times the common factor $\dot q_2$), do not cancel — the walls take up the difference, which is why a chain with fixed ends has no momentum conservation and no zero mode. Section 4 removes the walls and gets both back.

07 · the $M$-inner product · example

Two masses, two geometries

checks

Trace: $\lambda_1+\lambda_2 = \tfrac{10}{4} = \mathrm{tr}(M^{-1}K) = 2 + \tfrac12$. ✓

Determinant: $\lambda_1\lambda_2 = \tfrac34 = \det K/\det M = 3/4$. ✓

Limit: set $m_2 = m_1 = 1$ instead and the roots become $1$ and $3$, i.e. $\omega = \sqrt{k/m}$ and $\sqrt{3k/m}$ — the symmetric and antisymmetric modes of two equal masses, and the $N=2$ case of the chain's formula. ✓

The Euclidean angle between the modes is $67.4^\circ$; they are at right angles in the kinetic-energy geometry. Multiplying by $M^{1/2} = \mathrm{diag}(1,2)$ turns that geometry into the ordinary one.

Left: the modes, scaled to $\lVert v\rVert_M = 1$, end on the ellipse $u^{\!\top}\!Mu = u_1^2 + 4u_2^2 = 1$; $M$-orthogonal means the tangent at one is parallel to the other. Right: $w = M^{1/2}u$ stretches $u_2$ by $2$, the ellipse becomes the unit circle, and the same two vectors are perpendicular — the eigenvectors of $S$.

What to look for. On the left, the two arrows are not at right angles, and no rotation of the picture will make them so: the masses' own coordinates are the wrong geometry. But they are related to the ellipse $\lVert u\rVert_M = 1$ in a special way — each is parallel to the tangent at the other's tip. Such a pair is called a pair of conjugate diameters, and conjugacy is exactly $\langle v^{(1)}, v^{(2)}\rangle_M = 0$: the tangent to $u^{\!\top}\!Mu = 1$ at $v$ is the set of $x$ with $x^{\!\top}\!Mv = 0$.

On the right, the map $w = M^{1/2}u$ is applied to everything. A linear map carries ellipses to ellipses and tangents to tangents, and this one carries the $M$-unit ellipse to the Euclidean unit circle, where "the tangent at one is parallel to the other" means "perpendicular". The arrows on the right are, up to sign, the orthonormal eigenvectors of $S = M^{-1/2}KM^{-1/2}$, at $73.2^\circ$ and $-16.8^\circ$ from the $w_1$-axis — the same directions as the touching points in the spectral-theorem figure, which was drawn for this $S$.

The numbers. $\lVert v^{(1)}\rVert_M = \sqrt{1 + 4(1.651388)^2} = 3.450844$ and $\lVert v^{(2)}\rVert_M = \sqrt{1 + 4(0.151388)^2} = 1.044832$, so the drawn vectors are $(0.289784,\ 0.478546)$ and $(0.957092,\ -0.144892)$; their images under $\mathrm{diag}(1,2)$ have Euclidean dot product $0.277350 - 0.277350 = 0$.

07 · normal coordinates

Normal coordinates: the equations fall apart

Let $P = [\,v_1\ \cdots\ v_n\,]$ be the $M$-orthonormal modes as columns — the modal matrix. (It is called $P$, not $V$: $V$ is the potential energy.)

theorem · simultaneous diagonalisation

$$P^{\!\top}MP = I_n, \qquad P^{\!\top}KP = \mathrm{diag}(\omega_1^2,\dots,\omega_n^2).$$ Both forms are diagonal in the same basis — by congruence, not similarity.

Proof. Entry $(i,j)$ of the first is $\langle v_i,v_j\rangle_M = \delta_{ij}$; of the second, $v_i^{\!\top}Kv_j = \lambda_j v_i^{\!\top}Mv_j = \lambda_j\delta_{ij}$. $\square$

Writing $u(t) = \sum_p q_p(t)\,v_p$ defines the normal coordinates $q = P^{-1}u$, and substituting into $M\ddot u = -Ku$ then multiplying by $P^{\!\top}$ gives

$$\ddot q_p = -\omega_p^2\,q_p, \qquad p = 1,\dots,n .$$

one copy of unit 06 per mode

Each $q_p$ is an independent oscillator: $q_p(t) = \alpha_p\cos\omega_pt + \beta_p\sin\omega_pt$, with $\alpha_p = \langle u(0), v_p\rangle_M$ and $\beta_p = \langle\dot u(0), v_p\rangle_M/\omega_p$.

$2n$ constants, matching the count on “First order, in $2n$ dimensions”. ✓

and the energy too

$$T = \tfrac12\sum_p\dot q_p^2, \qquad V = \tfrac12\sum_p\omega_p^2q_p^2 .$$ No cross terms: the total energy is a sum of per-mode energies, each separately conserved. Modes do not exchange energy — in the linear approximation.

What the neglected $o(\lVert u\rVert^2)$ does to that last sentence is units 21 and 17: amplitude-dependent frequencies, then resonance between modes, and eventually chaos.

The substitution, in full. $u = Pq$ gives $MP\ddot q = -KPq$; left-multiply by $P^{\!\top}$ to get $(P^{\!\top}MP)\ddot q = -(P^{\!\top}KP)q$, i.e. $\ddot q = -\mathrm{diag}(\omega_p^2)q$. Note that $P^{\!\top}$, not $P^{-1}$, is what is applied — and it works because $P^{\!\top}MP = I$ means $P^{-1} = P^{\!\top}M$. The whole trick is that the transpose is the right inverse once $M$ is folded in.

Why the initial conditions come out as $M$-projections. From $u(0) = \sum_p\alpha_pv_p$, take $\langle\cdot, v_q\rangle_M$ of both sides: $\langle u(0),v_q\rangle_M = \sum_p\alpha_p\delta_{pq} = \alpha_q$. This is the standard formula for coefficients with respect to an orthonormal basis, in the $M$-inner product. Unit 19 will do the same computation with an integral in place of the sum, and it will look identical because it is the same argument.

On the phrase "normal". Read it as "orthogonal" — the modes are normal to one another, in the $M$-inner product — not as "typical" or "normalised". The name is nineteenth-century (Rayleigh's Theory of Sound, 1877, uses "normal coordinates" throughout); the mathematics is older, since Lagrange solved the loaded string mode by mode in 1759.

The energy statement, derived. $T = \tfrac12\dot u^{\!\top}M\dot u = \tfrac12\dot q^{\!\top}(P^{\!\top}MP)\dot q = \tfrac12\lVert\dot q\rVert^2$, and $V = \tfrac12 u^{\!\top}Ku = \tfrac12 q^{\!\top}\mathrm{diag}(\omega_p^2)q$. Each pair $E_p = \tfrac12(\dot q_p^2 + \omega_p^2q_p^2)$ is conserved separately, by unit 00's one-dimensional energy argument applied to $\ddot q_p = -\omega_p^2q_p$. A system of $n$ masses therefore has $n$ independent conserved quantities, not one — it is completely integrable, in the sense unit 17 will define.

07 · normal coordinates · example

One release, two clean oscillations

The two-mass example, released from rest at $u(0) = (1,\ 0.5)$. With the $M$-normalised modes of the last slides, the recipe $\alpha_p = \langle u(0), v_p\rangle_M$ gives $\alpha_1 = 1.2469$ and $\alpha_2 = 0.6673$, and $\beta_p = 0$ since nothing is moving. So

$$\begin{aligned}u_1(t) &= 0.3613\cos\omega_1t + 0.6387\cos\omega_2t,\\ u_2(t) &= 0.5967\cos\omega_1t - 0.0967\cos\omega_2t.\end{aligned}$$

Each coordinate is a mixture with incommensurable periods, so it never repeats. Each normal coordinate is one cosine.

Check at $t = 0$: $0.3613 + 0.6387 = 1$ and $0.5967 - 0.0967 = 0.5$. ✓

Top: what the masses do. Bottom: the same motion in normal coordinates, $q_p = \alpha_p\cos\omega_pt$ — the overview's "mess" taken apart. Computed from the closed form, not simulated.

Where the coefficients come from. With $v_1 = (0.289784,\ 0.478546)$ and $v_2 = (0.957092,\ -0.144892)$ (the modes scaled to $\lVert v\rVert_M = 1$, $M = \mathrm{diag}(1,4)$): $\alpha_1 = 1\cdot0.289784 + 4\cdot0.5\cdot0.478546 = 1.246876$ and $\alpha_2 = 0.957092 + 2\cdot(-0.144892) = 0.667308$. Then $u(t) = \sum_p\alpha_p\cos(\omega_pt)\,v_p$, and the coefficients on the slide are $\alpha_p$ times the components of $v_p$.

What to look for. In the top panel neither mass's motion has a period: the ratio $\omega_2/\omega_1 = 2.4842$ is irrational (its square, $\lambda_2/\lambda_1 = (19+5\sqrt{13})/6$, is), so the pattern never exactly recurs. The bottom panel is two cosines, one per mode, with the energy of each fixed for all time. Nothing in the top panel is noise or approximation; it is the bottom panel seen through the matrix $P$.

section 3

The chain, exactly

A sine that happens to be an eigenvector, a closed-form spectrum, and a curve bending away from a straight line.

07 · the chain

Writing down $M$ and $K$

$N$ masses $m$, $N+1$ springs $k$, fixed walls. Each of the $N+1$ springs stores $\tfrac12k\times(\text{its extension})^2$, and with the convention $u_0 = u_{N+1} = 0$ the walls are just two more values of $u$. So

$$V(u) = \frac k2\sum_{j=0}^{N}(u_{j+1}-u_j)^2 ,$$

and differentiating twice gives $K = kA_N$, $M = mI_N$, where

$$A_N = \begin{pmatrix} 2&-1&&\\ -1&2&\ddots&\\ &\ddots&\ddots&-1\\ &&-1&2\end{pmatrix}$$

is the second-difference operator: $(A_Nu)_j = -\big(u_{j+1} - 2u_j + u_{j-1}\big)$.

$K$ is positive definite

$u^{\!\top}A_Nu = \sum_{j=0}^{N}(u_{j+1}-u_j)^2 \ge 0$, and it vanishes only if all the differences vanish, i.e. $u_0 = u_1 = \dots = u_{N+1}$; the walls force that common value to be $0$.

So all $\omega_n > 0$ by the theorem on “What the reduction proves” — before computing anything. Take the walls away in section 4 and this argument is exactly what fails.

The equation of motion for mass $j$, read off directly from Hooke's law:

$$\begin{aligned}m\ddot u_j &= k(u_{j+1}-u_j) - k(u_j-u_{j-1})\\ &= k\,(u_{j+1}-2u_j+u_{j-1}) .\end{aligned}$$

Same thing. The Hessian route and the force route agree because the force is $-\nabla V$ (unit 04).

The Hessian, entry by entry. $\partial_i V = k\sum_j (u_{j+1}-u_j)(\delta_{i,j+1}-\delta_{i,j}) = k\big(2u_i - u_{i-1} - u_{i+1}\big)$, and differentiating again gives $\partial_l\partial_iV = k(2\delta_{il} - \delta_{i,l+1} - \delta_{i,l-1})$, which is $kA_N$. Positive semidefiniteness was guaranteed in advance because $u = 0$ minimises $V$, but the sum-of-squares form proves the stronger statement, definiteness, directly.

On the sign convention. $A_N$ is minus the discrete Laplacian, so that it is positive rather than negative definite. Books differ; a good habit is to fix the sign by demanding that the energy $\tfrac12u^{\!\top}Ku$ be positive, which is what was done here. Unit 18 will keep the same convention when $A_N/a^2$ becomes $-\partial_x^2$.

The second-difference operator is the whole of unit 18 in miniature. $(A_Nu)_j/a^2 = -\big(u_{j+1}-2u_j+u_{j-1}\big)/a^2$ is the standard second-difference quotient, which tends to $-u''(x)$ as $a \to 0$ for a smooth $u$ sampled at spacing $a$. The equation $m\ddot u_j = -k(A_Nu)_j$ therefore becomes $\rho\,\partial_t^2 u = \tau\,\partial_x^2u$ in the limit, the wave equation. That limit is taken properly in unit 18; “A continuum theory's own error bar” measures how far the finite chain is from it.

07 · the chain · example

Worked: a sine is an eigenvector

worked · the ansatz, verified

1. Try $v_j = \sin(j\theta)$, $j = 0,\dots,N+1$, $\theta$ to be chosen.

2. Apply $A_N$. The identity $\sin\big((j{-}1)\theta\big) + \sin\big((j{+}1)\theta\big) = 2\cos\theta\,\sin(j\theta)$ (add the two angle-sum formulas) gives, for every $j$, $$-v_{j-1} + 2v_j - v_{j+1} = (2-2\cos\theta)\,v_j .$$

3. The walls. $v_0 = \sin 0 = 0$ automatically. The other end needs $v_{N+1} = \sin\big((N{+}1)\theta\big) = 0$, so $\theta_n = n\pi/(N{+}1)$ with $n \in \mathbb Z$.

4. Which $n$ are new. Every other integer repeats one of $n = 1,\dots,N$ up to sign, or gives $0$ (notes). That is $N$ modes — all of them, by the count on “What the reduction proves”.

5. Frequencies. $\lambda_n = \tfrac km(2-2\cos\theta_n) = \tfrac{4k}m\sin^2(\theta_n/2)$, so

$$\boxed{\ \omega_n = 2\sqrt{\frac km}\,\sin\frac{n\pi}{2(N+1)}\ }$$

Step 2 holds at $j = 1$ and $j = N$ too: the walls are $v_0 = v_{N+1} = 0$, built into the ansatz.

$N = 5$: each mass's displacement is the height of a sine at that mass (dashed), and the sine vanishes at both walls: $(0.5,\ 0.866,\ 1,\ 0.866,\ 0.5)$ and $(0.866,\ 0.866,\ 0,\ -0.866,\ -0.866)$. Mode $n$ has $n-1$ interior nodes.

The trigonometric identity, spelt out. $\sin((j-1)\theta) = \sin j\theta\cos\theta - \cos j\theta\sin\theta$ and $\sin((j+1)\theta) = \sin j\theta\cos\theta + \cos j\theta\sin\theta$; adding kills the cosine terms and leaves $2\cos\theta\sin j\theta$. The complex-exponential route of unit 06 is quicker: $v_j = \mathrm{Im}\,z^j$ with $z = e^{i\theta}$, and $z^{-1} + z = 2\cos\theta$ is immediate.

Why a sine, and not a guess out of nowhere. The matrix $A_N$ commutes with the reflection $j \mapsto N+1-j$, so its eigenvectors can be taken even or odd about the centre; and the recursion $v_{j+1} = (2-\lambda m/k)v_j - v_{j-1}$ is a linear difference equation with constant coefficients, whose solutions are $z^j$ for $z$ a root of $z^2 - (2-\lambda m/k)z + 1 = 0$. The two roots multiply to $1$, so either both are real reciprocals — giving growing and decaying solutions that cannot satisfy both boundary conditions — or they are $e^{\pm i\theta}$ on the unit circle, giving $\sin j\theta$ and $\cos j\theta$. The ansatz is the general solution of a constant-coefficient recursion, exactly as in unit 06 for a constant-coefficient ODE.

Step 4, carefully. $n = 0$ and $n = N+1$ give $\sin(0) = 0$ and $\sin(j\pi) = 0$ for every $j$, the zero vector; $-n$ gives minus the vector of $n$, since sine is odd; and $\sin\big(j(2(N{+}1)-n)\pi/(N{+}1)\big) = \sin\big(2\pi j - jn\pi/(N{+}1)\big) = -\sin\big(jn\pi/(N{+}1)\big)$, so those two indices give parallel vectors. Together with periodicity in $n$ with period $2(N{+}1)$, every integer $n$ reduces to one of $1,\dots,N$ or to zero. Since “What the reduction proves” promised exactly $N$ modes and we have found $N$ independent ones — independent because their eigenvalues $4\sin^2\big(n\pi/2(N{+}1)\big)$ are distinct, the sine being strictly increasing on $[0,\pi/2]$ — the list is complete — no separate completeness argument is needed.

Node counting. $v^{(n)}_j = \sin(jn\pi/(N+1))$ changes sign $n-1$ times between the walls. That mode $n$ has $n-1$ nodes is a general fact about second-order problems (Sturm's oscillation theorem, unit 19); here it can simply be read off.

07 · the chain

Four checks on the formula

dimensions

$\sqrt{k/m} = \sqrt{(\mathrm{N\,m^{-1}})/\mathrm{kg}} = \mathrm{s^{-1}}$, and the sine's argument is a pure number. ✓

$N = 1$

$\omega_1 = 2\sqrt{k/m}\sin(\pi/4) = \sqrt{2k/m}$. Direct: one mass between two springs feels restoring force $2ku$, so $\omega = \sqrt{2k/m}$. ✓

$N = 2$

$\omega_1 = 2\sin(\pi/6)\sqrt{k/m} = \sqrt{k/m}$: both masses together, middle spring unstretched, each feels only its wall spring. ✓
$\omega_2 = 2\sin(\pi/3)\sqrt{k/m} = \sqrt{3k/m}$: opposite motion stretches the middle spring by $2u$, giving effective stiffness $k + 2k = 3k$. ✓

the ceiling

$\omega_n < 2\sqrt{k/m}$ always, approached only as $\theta \to \pi$: neighbours in exact antiphase. However many masses you add, nothing vibrates faster — the cut-off frequency.

For the $N = 20$ chain: ceiling $31.623\ \mathrm{s^{-1}}$, top mode $\omega_{20} = 31.534\ \mathrm{s^{-1}}$, which is $99.72\%$ of it.

orthogonality, as promised

$M = mI$ here, so $M$-orthogonality is Euclidean orthogonality, and the theorem hands us the vanishing of $$\sum_{j=1}^{N}\sin\frac{jn\pi}{N{+}1}\,\sin\frac{jn'\pi}{N{+}1} \quad (n \ne n'\ \text{in}\ 1,\dots,N),$$ a discrete Fourier identity, proved by physics; the diagonal value is $(N{+}1)/2$ (notes).

The $N=2$ check, done as a pencil problem. $M = mI_2$, $K = k\begin{pmatrix}2&-1\\-1&2\end{pmatrix}$, so $\det(K-\lambda mI) = (2k-\lambda m)^2 - k^2$, giving $\lambda m = k$ or $3k$ — exactly the two frequencies above, and the eigenvectors $(1,1)$ and $(1,-1)$ are the symmetric and antisymmetric shapes. They are Euclidean-orthogonal because the masses are equal; the worked two-mass example is what happens when they are not.

Why the ceiling is physical, not an artefact. The stiffest thing a single mass can experience is both its springs pulling it back at once, with its neighbours displaced the opposite way: restoring force $k(u - (-u)) \times 2 = 4ku$, hence $\omega^2 = 4k/m$. No arrangement does better, so $2\sqrt{k/m}$ bounds the spectrum. Section 5's Rayleigh argument proves exactly this, from above: $\lambda_{\max} = \max R(u)$ and $R(u) \le 4k/m$ because $\sum(u_{j+1}-u_j)^2 \le 2\sum(u_{j+1}^2+u_j^2) \le 4\sum u_j^2$.

The discrete orthogonality identity. It can also be proved directly, by writing the sines as exponentials and summing two geometric series; the physics proof above is shorter and says why it is true. The normalisation $(N+1)/2$ follows from $\sum_j\sin^2 = \tfrac12\sum_j(1-\cos(2jn\pi/(N{+}1))) = \tfrac N2 + \tfrac12$, using that the cosine sum is $-1$ for $1 \le n \le N$. So the $M$-orthonormal modes are $v^{(n)}_j = \sqrt{2/\big(m(N{+}1)\big)}\,\sin\big(jn\pi/(N{+}1)\big)$, with the $\sqrt m$ there because $\langle v,v\rangle_M = m\lVert v\rVert^2$.

07 · the chain

The dispersion curve, and where it bends

Give the chain a spacing $a$ and define the wavenumber $\kappa_n = \theta_n/a = n\pi/\big((N{+}1)a\big)$, so a mode looks like $\sin(\kappa_n x_j)$ with $x_j = ja$. Then

$$\omega(\kappa) = \frac{2c}{a}\,\sin\frac{\kappa a}{2}, \qquad c = a\sqrt{k/m},$$

the chain's dispersion relation. For $\kappa a \ll 1$ it is $\omega \approx c\kappa$: a straight line, one speed for all wavelengths. For $\kappa a$ of order $1$ it is not.

$n$	$\kappa_n a$	$\omega_n/\sqrt{k/m}$	$c\kappa_n$	deficit
1	0.14960	0.14946	0.14960	0.093%
5	0.74800	0.73068	0.74800	2.32%
10	1.49600	1.36034	1.49600	9.07%
20	2.99199	1.99441	2.99199	33.34%

$N = 20$, in units of $\sqrt{k/m}$ with $a = 1$.

Dots: the twenty $\omega_n$ for $N = 20$, on the arch $2\sin(\kappa a/2)$. Dashed: the straight line $c\kappa$ it is tangent to at the origin. The gap between them is what unit 18's continuum theory does not know about.

All five columns were computed and checked against a direct numerical eigendecomposition of $A_{20}$: the largest discrepancy between the closed formula and the numerical eigenvalues is of order $10^{-15}$, which is rounding.

What "dispersion" means, and why it matters. If $\omega = c\kappa$ exactly, every wavelength travels at the same speed $c$ and a pulse keeps its shape forever. When $\omega(\kappa)$ bends, short wavelengths travel slower than long ones — here the phase speed is $\omega/\kappa = c\,\mathrm{sinc}(\kappa a/2) < c$ — so a pulse made of many wavelengths spreads out as it travels. A crystal lattice is exactly this chain in three dimensions, and its dispersion curve, measured by neutron scattering, is one of the standard confirmations that matter is made of discrete atoms on a lattice.

The top of the curve is flat. $\mathrm d\omega/\mathrm d\kappa = c\cos(\kappa a/2)$ vanishes at $\kappa a = \pi$. A wave packet built near the cut-off does not travel at all: its group velocity is zero. That $\kappa a = \pi$ is the edge of what the lattice can represent — the shortest meaningful wavelength is $2a$, one atom up and the next down — and in solid-state physics it is the edge of the first Brillouin zone. The same fact is the Nyquist limit in signal processing.

07 · the chain

A continuum theory's own error bar

Expand the exact relation for small $\kappa a$, using $\sin x = x - x^3/6 + x^5/120 - \dots$ with $x = \kappa a/2$:

the correction $$\omega(\kappa) = c\kappa\Big(1 - \frac{(\kappa a)^2}{24} + \frac{(\kappa a)^4}{1920} - \dots\Big).$$

Proof. $\omega = (2c/a)\sin(\kappa a/2) = c\kappa\cdot\dfrac{\sin(\kappa a/2)}{\kappa a/2}$, and $\sin x/x = 1 - x^2/6 + x^4/120$ with $x = \kappa a/2$. $\square$

Unit 18 will derive the wave equation as the $a \to 0$ limit and predict $\omega = c\kappa$ exactly. The $-(\kappa a)^2/24$ is the size of the lie, and it is computable in advance.

worked · the fundamental of the $N=20$ chain

Continuum prediction. A string of length $L = 1.05\ \mathrm m$ with wave speed $c = a\sqrt{k/m} = 0.05\times15.8114 = 0.79057\ \mathrm{m\,s^{-1}}$ has fundamental $\omega = \pi c/L = 2.36538\ \mathrm{s^{-1}}$.

The chain. $\omega_1 = 2\sqrt{k/m}\sin(\pi/42) = 2\times15.8114\times0.0747300 = 2.36317\ \mathrm{s^{-1}}$.

The gap. $(2.36317 - 2.36538)/2.36538 = -0.0932\%$.

Predicted gap. $\kappa_1a = \pi/21 = 0.149600$, so $-(\kappa_1a)^2/24 = -0.0933\%$.

Three significant figures, from a Taylor series, with no fitting.

The next slide shows the whole chain, mode by mode.

This slide is the one to remember from section 3, and not for the chain. It is a working example of something rare: a continuum theory whose error, relative to the discrete theory underneath it, is known in closed form. Usually one has the continuum theory and no access at all to what it is approximating.

Where the $1/24$ comes from, physically. Replacing $u_{j+1}-2u_j+u_{j-1}$ by $a^2u''(x_j)$ throws away $\tfrac{a^4}{12}u''''(x_j) + O(a^6)$ — Taylor's theorem again. For $u = \sin\kappa x$ that discarded term is $\tfrac{a^4\kappa^4}{12}\sin\kappa x$, so the corrected eigenvalue is $\omega^2 = c^2\kappa^2\big(1 - (\kappa a)^2/12\big)$, and taking the square root halves the exponent's coefficient: $1 - (\kappa a)^2/24$. The two routes agree, as they must.

Why the deficit is always negative. $\sin x < x$ for $x > 0$, so $\omega(\kappa) < c\kappa$ for every $\kappa > 0$: the discrete chain is always softer than the continuum string predicts. That is a fact about this discretisation — point masses joined by springs — and not about discretisation in general. Section 5 shows that a Rayleigh–Ritz model, which minimises the continuum's own energy ratio over a restricted family of shapes, always errs high. Which side a finite model lands on depends on how it was built, and the chain was not built that way.

07 · the chain

The deficit, mode by mode

The exact relative deficit of mode $n$ below the continuum line is $$1 - \frac{\sin(\kappa_na/2)}{\kappa_na/2}, \qquad \kappa_na = \frac{n\pi}{21},$$ and the one-term prediction is $(\kappa_na)^2/24$.

At $n = 5$ the prediction is $2.331\%$ against an actual $2.315\%$; at $n = 10$, $9.33\%$ against $9.07\%$; at $n = 20$, $37.3\%$ against $33.3\%$.

as a truncated series should

Excellent where $\kappa a$ is small, degrading smoothly where it is not, and always on the same side: the next term, $+(\kappa a)^4/1920$, is positive, so the prediction overstates the deficit. Unit 21 makes that sentence precise.

Both axes logarithmic. Dots: the exact deficit of each of the twenty modes. Dashed: $(\kappa a)^2/24$, a power law, so a straight line of slope $2$. The dots sit on it for small $n$ and fall below it as $\kappa a$ nears $\pi$.

Every number, recomputed. The deficits $1 - \mathrm{sinc}$ at $n = 1, 5, 10, 20$ are $0.0932\%$, $2.315\%$, $9.07\%$ and $33.34\%$, as in the table on “The dispersion curve, and where it bends”. The predictions $(n\pi/21)^2/24$ are $0.0933\%$, $2.331\%$, $9.33\%$ and $37.30\%$. With the next term included, $(\kappa a)^2/24 - (\kappa a)^4/1920$, the $n = 20$ prediction becomes $33.13\%$ — the alternating series closing in from alternate sides.

Why the dashed curve lies above. $\sin x/x = 1 - x^2/6 + x^4/120 - \cdots$ is an alternating series whose terms decrease in size for $0 < x < \pi/2$ (here $x = \kappa a/2 \le \pi/2$), so the truncation error has the sign of the first omitted term and is smaller than it. Stopping after $-x^2/6$ omits $+x^4/120$, so $\sin x/x$ is larger than $1 - x^2/6$, and the true deficit smaller than predicted.

section 4

Zero modes, and the symmetries responsible

Take the walls away and one frequency goes to zero. That is not an accident and not a degeneracy: it is a symmetry, and behind it a conservation law — six units before Noether (unit 13) says why.

07 · zero modes

Cut the chain loose

Remove the two walls from the chain: $N$ masses, $N-1$ springs, nothing fixed. Now $V = \tfrac k2\sum_{j=1}^{N-1}(u_{j+1}-u_j)^2$, and $K = k\tilde A_N$ with $\tilde A_N$ equal to $A_N$ except that the two corner entries are $1$ instead of $2$.

the spectrum, free ends

$$\omega_n = 2\sqrt{\frac km}\,\sin\frac{n\pi}{2N}, \qquad n = 0,1,\dots,N-1,$$ so $\omega_0 = 0$ with mode $v^{(0)} = (1,1,\dots,1)$.

Check it on $N = 3$: the next slide.

definition

A zero mode is a solution of $Kv = 0$, $v \ne 0$: a displacement direction that costs no potential energy to second order.

why the old argument failed

“Writing down $M$ and $K$” proved $K$ positive definite by saying the walls force the common value $u_0 = \dots = u_{N+1}$ to be zero. With no walls, $u = (1,1,\dots,1)$ has every difference zero: $u^{\!\top}Ku = 0$ with $u \ne 0$. Positive semidefinite, not definite.

Verifying the free-end formula. The corner entry $1$ instead of $2$ is exactly the statement that the missing end spring has zero extension, so introduce ghost values $v_0 = v_1$ and $v_{N+1} = v_N$; with those, $(\tilde A_Nv)_j = -v_{j-1}+2v_j-v_{j+1}$ for every $j = 1,\dots,N$, the same expression as in the fixed case. Now try $v_j = \cos\big((j-\tfrac12)\theta\big)$. The left ghost condition holds automatically, since $\cos\big(-\tfrac{\theta}{2}\big) = \cos\big(\tfrac{\theta}{2}\big)$. The right one, $\cos\big((N+\tfrac12)\theta\big) = \cos\big((N-\tfrac12)\theta\big)$, is $-2\sin(N\theta)\sin(\theta/2) = 0$ by the difference-of-cosines identity, hence $\sin(N\theta) = 0$ and $\theta = n\pi/N$ for $n = 0,\dots,N-1$. The angle-sum identity then gives $(\tilde A_Nv)_j = (2-2\cos\theta)v_j$, so $\lambda_n = (k/m)(2-2\cos\theta_n) = (4k/m)\sin^2(n\pi/2N)$, which is the stated $\omega_n$. At $n = 0$ the vector is $(1,1,\dots,1)$.

The numerical check. Diagonalising $\tilde A_N$ for $N = 3, 4, 6$ returns exactly $\{0,\,1,\,1.7320508\}$, $\{0,\,0.7653669,\,1.4142136,\,1.8477591\}$ and $\{0,\,0.5176381,\,1,\,1.4142136,\,1.7320508,\,1.9318517\}$ in units of $\sqrt{k/m}$, agreeing with $2\sin(n\pi/2N)$ to machine precision, and the null vector comes back as the constant vector each time.

A curiosity worth noticing. The free chain of $N$ masses has the same nonzero frequencies as the fixed chain of $N-1$ masses. That is not a coincidence: reflecting a free-end solution about the end produces a fixed-end solution on the doubled chain, and the arithmetic $2\sin(n\pi/2N)$ against $2\sin(n\pi/2N)$ makes it visible. Unit 19 calls this the relation between Neumann and Dirichlet boundary conditions.

07 · zero modes

The free chain's three modes

$N = 3$, no walls: $K = k\begin{pmatrix}1&-1&0\\-1&2&-1\\0&-1&1\end{pmatrix}$, $M = mI$. The formula gives $\omega = 0,\ \sqrt{k/m},\ \sqrt{3k/m}$.

$(1,1,1)$: $K$ kills it — each row sums to $0$. Sliding the whole chain sideways costs nothing, because nothing is attached to anything fixed.
$(1,0,-1)$: the middle mass is stationary, so this is exactly two independent mass–spring pairs, $\omega = \sqrt{k/m}$. ✓
$(1,-2,1)$: $K$ gives $k(3,-6,3)$, three times the shape, so $\omega^2 = 3k/m$. ✓

Verified numerically for $N = 3, 4, 6$ against the formula. The three shapes are orthogonal, as $M = mI$ requires.

The three modes of three free masses, as displacement arrows. The top one is not a vibration at all: nothing restores it, so nothing makes it oscillate.

What to look for. In the two lower rows the arrows add up to zero, $1 + 0 - 1 = 0$ and $1 - 2 + 1 = 0$: those modes do not move the centre of mass. That is $M$-orthogonality to the zero mode, $\langle v, (1,1,1)\rangle_M = m\sum_jv_j = 0$, read as physics: every genuine vibration of a free body leaves its centre of mass where it was, and all the motion of the centre of mass is in the zero mode. “What a zero mode actually does” follows that motion.

The middle row, from the formula. The free-end modes are $v_j = \cos\big((j-\tfrac12)n\pi/N\big)$; for $N = 3$, $n = 1$ this is $(\cos 30^\circ, \cos 90^\circ, \cos 150^\circ) = (0.866, 0, -0.866)$, proportional to $(1,0,-1)$, and $n = 2$ gives $(\cos 60^\circ, \cos 180^\circ, \cos 300^\circ) = (0.5, -1, 0.5)$, proportional to $(1,-2,1)$.

07 · zero modes

A symmetry of $V$ is a null vector of $K$

theorem · a symmetry is a null vector, six units before Noether

Let $u^*$ be an equilibrium of $V$ and let $s \mapsto \varphi(s)$ be a $C^2$ curve in configuration space with $\varphi(0) = u^*$ and $V(\varphi(s)) = V(u^*)$ for all $s$ near $0$. Then $w = \varphi'(0)$ satisfies $$w^{\!\top}Kw = 0, \quad\text{and hence}\quad Kw = 0 .$$

proof

$g(s) = V(\varphi(s))$ is constant, so $g''(0) = 0$. The chain rule (unit 03) gives $g'(s) = \nabla V(\varphi(s))\cdot\varphi'(s)$ and $$g''(0) = \varphi'(0)^{\!\top}\mathrm{Hess}\,V(u^*)\,\varphi'(0) + \nabla V(u^*)\cdot\varphi''(0),$$ whose second term vanishes at an equilibrium. So $w^{\!\top}Kw = 0$, and Cauchy–Schwarz for the semidefinite form $K$ upgrades that to $Kw = 0$. $\square$

the Cauchy–Schwarz step in full, in the notes (N)

A point in the plane in a round trough, $V(u) = (\lVert u\rVert - 1)^2$: contours of $V$, and the floor, a circle of equilibria. Rotating $u^*$ keeps $V$ fixed, so the tangent $w$ is a zero mode; the radial direction is stiff. The curve bends away from $w$, but $\varphi''(0)$ meets $\nabla V(u^*) = 0$ and drops out.

What to look for in the figure. Here $K$ is the Hessian of $(\lVert u\rVert-1)^2$ at a point of the unit circle, which is $2\hat r\hat r^{\!\top}$ with $\hat r = u^*$: a stiffness $2$ along the radius and $0$ along the tangent. The tangent is the zero mode, and it is there because rotating $u^*$ about the centre keeps it on the floor. This is the two-dimensional picture of a molecule's rotational zero modes: the symmetry orbit is curved, its tangent is still a null vector.

The semi-inner product step, in full. For positive semidefinite $K$, the form $\langle x,y\rangle_K = x^{\!\top}Ky$ is symmetric, bilinear and satisfies $\langle x,x\rangle_K \ge 0$, which is all Cauchy–Schwarz needs: for real $t$, $0 \le \langle w+ty, w+ty\rangle_K = \langle w,w\rangle_K + 2t\langle w,y\rangle_K + t^2\langle y,y\rangle_K$, a quadratic in $t$ that is nowhere negative, so its discriminant is $\le 0$: $\langle w,y\rangle_K^2 \le \langle w,w\rangle_K\langle y,y\rangle_K$. With $\langle w,w\rangle_K = 0$ this forces $\langle w,y\rangle_K = 0$ for all $y$, i.e. $Kw = 0$. Definiteness is not needed, and that matters: $K$ is exactly the case where it fails.

The converse is false in general. A zero mode need not come from a symmetry of $V$; it can also be an accidental degeneracy of the quadratic approximation, with the motion actually restrained at fourth order. The standard example is $V = u_1^4 + u_2^2$, whose Hessian at the origin has a zero eigenvalue but whose potential is a genuine minimum in that direction. Such a mode is called soft rather than free, and its frequency is amplitude-dependent — which is a nonlinear effect, and out of scope until unit 21.

07 · zero modes · example

Counting zero modes

worked · counting zero modes

Free chain. $\varphi(s) = u^* + s(1,\dots,1)$ slides everything along the line; $V$ depends only on differences, so it is invariant. $w = (1,\dots,1)$: the zero mode found on “Cut the chain loose”, now explained.

A molecule in space, $n$ atoms, $3n$ coordinates, $V$ invariant under translations and rotations. Three translations give three zero modes; three rotations $\varphi(s) = R(s)u^*$ give $w = \Omega u^*$ with $\Omega \in \mathfrak{so}(3)$ (unit 03) — three more, or two if the molecule is straight.

Count. $3n - 6$ vibrational modes, or $3n - 5$ if linear. For $\mathrm{CO_2}$ ($n = 3$, linear): $\mathbf 4$.

Unit 13 turns this into Noether's theorem proper: there a symmetry gives a conserved quantity, here a symmetry gives a direction of zero restoring force. Same hypothesis, two faces.

$\mathrm{CO_2}$'s four vibrations: $3\times3 = 9$ coordinates, minus three translations and two rotations (spinning about its own axis moves no atom). The two bends are the same motion turned through $90^\circ$ about the axis, so they share a frequency. Schematic.

On the CO₂ number. Carbon dioxide is linear (O=C=O), so $3\times3-5 = 4$ vibrational modes: a symmetric stretch, an antisymmetric stretch, and two degenerate bends at right angles to each other. The degeneracy of the two bends is forced by the rotational symmetry about the molecular axis, which is the general rule mentioned on “What the reduction proves” — degeneracies are the fingerprints of symmetry groups. The symmetric stretch does not change the dipole moment and so is infrared-inactive, and the two bends share one frequency, which is why the infrared spectrum shows two fundamental bands and not four: the bend near $667\ \mathrm{cm^{-1}}$ and the antisymmetric stretch near $2349\ \mathrm{cm^{-1}}$ are the ones that make it a greenhouse gas.

07 · zero modes

What a zero mode actually does

In normal coordinates the zero mode obeys $\ddot q_0 = 0$, so

$$q_0(t) = \alpha + \beta t .$$

Not an oscillation: a uniform drift. For the free chain that is the centre of mass moving at constant velocity, which is momentum conservation — and $\beta = \langle\dot u(0), v^{(0)}\rangle_M = \sum_j m\dot u_j(0)$ is the total momentum, up to normalisation.

the Jordan block returns

In the $2n$-dimensional system $\dot y = Ay$, a zero mode contributes eigenvalue $0$ with algebraic multiplicity $2$ and only one eigenvector: a genuine $2\times2$ Jordan block, whose exponential $\begin{pmatrix}1&t\\0&1\end{pmatrix}$ is the $\alpha+\beta t$ above.

This is the only way a Jordan block can occur in an undamped system. Section 1 was not scenery.

Three free masses, the first kicked at $t = 0$ (rest spacing $a$, unit kick speed $a\sqrt{k/m}$). Every mass wiggles — the two vibrations — and all of them drift: the zero mode. Their mean, the centre of mass, moves on a straight line of slope $\tfrac13$.

The Jordan block, explicitly. Restricted to the zero mode's two-dimensional invariant subspace $\{(\alpha v^{(0)}, \beta v^{(0)})\}$ of $\mathbb R^{2n}$, the matrix $A = \begin{pmatrix}0&I\\-M^{-1}K&0\end{pmatrix}$ acts as $\begin{pmatrix}0&1\\0&0\end{pmatrix}$, since $Kv^{(0)} = 0$ kills the bottom-left entry. That is the nilpotent $N$ of “The degenerate case” with $N^2 = 0$, and that slide's theorem gives $e^{At} = I + Nt = \begin{pmatrix}1&t\\0&1\end{pmatrix}$. The free particle from the limit check on “When $A$ is diagonalisable”, one more time.

The figure, computed. With $k = m = a = 1$ and modes $v^{(0)} = (1,1,1)/\sqrt3$, $v^{(1)} = (1,0,-1)/\sqrt2$, $v^{(2)} = (1,-2,1)/\sqrt6$ at $\omega = 0, 1, \sqrt3$, the kick $\dot u(0) = (1,0,0)$ gives $q_0 = t/\sqrt3$, $q_1 = \sin t/\sqrt2$, $q_2 = \sin(\sqrt3\,t)/(\sqrt6\sqrt3)$, and the drawn positions are $j + u_j(t)$ with $u = \sum_p q_pv^{(p)}$. The zero-mode part of every mass is $q_0/\sqrt3 = t/3$: total momentum $1$ shared by total mass $3$.

A practical warning. A numerical diagonalisation returns a zero mode as something like $10^{-16}$, sometimes slightly negative, and $\sqrt{\lambda}$ then comes back as a tiny imaginary number. That is rounding, not physics. The honest check is whether the corresponding eigenvector is the symmetry direction the problem obviously has — and if it is, set the frequency to zero by hand.

07 · zero modes

The three regimes, side by side

the three regimes, complete

$\lambda = \omega^2$	motion	meaning
$> 0$	$\cos\omega t,\ \sin\omega t$	stable direction
$= 0$	$\alpha + \beta t$	symmetry: a zero mode
$< 0$	$e^{\pm\|\omega\|t}$	unstable; unit 02's saddle

No fourth case. In particular no $\omega$ with both parts nonzero — growing oscillation needs damping or driving, and neither is in $M\ddot u = -Ku$ with $M, K$ symmetric.

That the list is complete is the theorem of “What the reduction proves”: $\lambda$ real, always; a full basis of modes, always.

One normal coordinate in its phase plane $(q, \dot q)$, for each sign of $\lambda$. Closed orbits; straight lines along which $q$ drifts at constant $\dot q$, with a whole axis of rest points; a saddle, two directions in and two out. Nothing spirals: a spiral is the forbidden complex $\omega$.

What to look for. The middle panel is the Jordan block of “What a zero mode actually does”: the flow $e^{At}$ with $A = \begin{pmatrix}0&1\\0&0\end{pmatrix}$ shears the plane, sliding each horizontal line at its own speed $\dot q$. The other two panels are its neighbours: perturb $\lambda$ from $0$ to a small positive value and the lines close up into very long thin ellipses; to a small negative value and they open into hyperbolas. All three have a conserved quantity, $\tfrac12\dot q^2 + \tfrac12\lambda q^2$, and the pictures are its level sets.

Why "growing oscillation" is forbidden, again but structurally. The energy $E = \tfrac12\dot u^{\!\top}M\dot u + \tfrac12u^{\!\top}Ku$ satisfies $\dot E = \dot u^{\!\top}M\ddot u + \dot u^{\!\top}Ku = \dot u^{\!\top}(-Ku) + \dot u^{\!\top}Ku = 0$, where symmetry of $K$ is used to write the derivative of the quadratic form that way. If $K$ is positive definite, $E$ is a positive definite quadratic function of $(u,\dot u)$ that is conserved, so every trajectory stays on a bounded level set: nothing can grow, ever. That is Lyapunov's argument from unit 02, with $E$ as the Lyapunov function, and it gives the conclusion without any eigenvalue computation at all.

section 5

Rayleigh: bounding a frequency without solving

The lowest frequency is the minimum of a ratio. So any guess at the shape, however crude, is an upper bound — and you know the direction of the error.

07 · Rayleigh

The Rayleigh quotient

definition

For $u \ne 0$, the Rayleigh quotient of the pair $(K, M)$ is $$R(u) = \frac{u^{\!\top}Ku}{u^{\!\top}Mu} = \frac{2V(u)}{\lVert u\rVert_M^2} .$$ It is scale-invariant: $R(cu) = R(u)$, so it is really a function on directions.

theorem · the lowest frequency is a minimum

$\omega_1^2 = \lambda_1 = \min_{u \ne 0} R(u)$, attained exactly on the lowest mode's eigenspace; and $\lambda_n = \max_u R(u)$.

Proof. Expand $u = \sum_p c_pv_p$ in the $M$-orthonormal modes. Then $u^{\!\top}Mu = \sum_pc_p^2$ and $u^{\!\top}Ku = \sum_p\lambda_pc_p^2$, so $R(u) = \sum_p\lambda_pc_p^2\big/\sum_pc_p^2$ is a weighted average of $\lambda_1,\dots,\lambda_n$ with nonnegative weights. Hence $\lambda_1 \le R(u) \le \lambda_n$, with equality on the left iff $c_p = 0$ whenever $\lambda_p > \lambda_1$. $\square$

The physics. $R(u)$ is twice the potential energy of the shape $u$ over twice the kinetic energy of moving with that shape at unit rate. The system vibrates in whatever shape makes the ratio smallest: stiffness cheap, inertia dear.

The spectral theorem's proof was this argument already: its eigenvector was a maximiser of the same kind of quotient. The next slide draws $R$ for the two-mass example.

Full proof of the min–max (Courant–Fischer) characterisation

The statement is $\displaystyle\lambda_p = \min_{\dim W = p}\ \max_{u\in W\setminus 0}R(u)$, over subspaces $W \subseteq \mathbb R^n$. Upper bound: take $W = \mathrm{span}(v_1,\dots,v_p)$; on it $R$ is a weighted average of $\lambda_1,\dots,\lambda_p$, so its maximum is $\lambda_p$, and the min over $W$ is at most $\lambda_p$. Lower bound: let $W$ be any $p$-dimensional subspace and $U = \mathrm{span}(v_p,\dots,v_n)$, of dimension $n-p+1$. Then $\dim(W\cap U) \ge p + (n-p+1) - n = 1$, so there is a nonzero $u \in W\cap U$; on $U$, $R(u)$ is a weighted average of $\lambda_p,\dots,\lambda_n$, hence $\ge\lambda_p$. So $\max_{W}R \ge \lambda_p$ for every $W$. $\square$

Two consequences worth knowing. (i) Monotonicity: if $K$ is replaced by $K' \succeq K$ (stiffening: $u^{\!\top}K'u \ge u^{\!\top}Ku$ for all $u$) then every $\lambda_p$ can only increase, since $R$ increases pointwise. Adding mass lowers them, for the same reason applied to the denominator. (ii) Interlacing: imposing one linear constraint — clamping a point — restricts the minimisation to a hyperplane, and the new frequencies interlace the old: $\lambda_p \le \lambda'_p \le \lambda_{p+1}$. Both follow from min–max in one line each, and both are exactly what an engineer's intuition says.

07 · Rayleigh

The quotient for two masses, drawn

For the two-mass example, $R$ depends only on the direction of $u = (\cos\varphi, \sin\varphi)$: $$R(\varphi) = \frac{2 - 2\cos\varphi\sin\varphi}{\cos^2\varphi + 4\sin^2\varphi}\ \mathrm{s^{-2}}.$$

what it is good for

Any trial vector bounds $\omega_1$ from above — no solving, and the error is one-sided. And the bound is forgiving: at the minimum $R$ has zero gradient, so an error $\varepsilon$ in the shape costs only $O(\varepsilon^2)$ in the frequency.

On the plot: $15^\circ$ off the true shape, $R = 0.3888$ against $\lambda_1 = 0.3486$, so $\omega_1 \le 0.6235$ against the true $0.5904$ — $5.6\%$ high, from a $15^\circ$ ($0.26$ rad) error in the shape.

$R$ over every direction in the plane. Its minimum is $\lambda_1$, at the direction of $v^{(1)}$; its maximum is $\lambda_2$, at $v^{(2)}$; every other value lies between. The bottom is flat: that is the quadratic insensitivity.

The quadratic insensitivity, proved. Write the trial vector as $u = v_1 + \varepsilon z$ with $\langle z, v_1\rangle_M = 0$ and $\lVert z\rVert_M = 1$. Expanding $z = \sum_{p\ge2}c_pv_p$, $$R(u) = \frac{\lambda_1 + \varepsilon^2\sum_{p\ge2}\lambda_pc_p^2}{1+\varepsilon^2} = \lambda_1 + \varepsilon^2\sum_{p\ge2}(\lambda_p - \lambda_1)c_p^2 + O(\varepsilon^4).$$ So the relative error in $\lambda_1$ is second order in the shape error — which is why Rayleigh's method was a practical engineering tool for a century before computers, and why finite-element frequencies converge quickly.

The figure's numbers. $v^{(1)} = (1,\ 1.651388)$ points at $\varphi = \arctan 1.651388 = 58.80^\circ$, and $R$ there is $0.348612$; $v^{(2)} = (1,\ -0.151388)$ points at $180^\circ - 8.61^\circ = 171.39^\circ$, where $R = 2.151388$. The point drawn at $73.80^\circ$ has $R = 0.388751$.

07 · Rayleigh · example

Worked: bounding the fundamental of a five-mass chain

worked · the five-mass chain

The quotient. With $M = mI$, $$R(u) = \frac km\cdot\frac{\sum_{j=0}^{5}(u_{j+1}-u_j)^2}{\sum_{j=1}^{5}u_j^2}, \qquad u_0 = u_6 = 0 .$$

Trial 1 · flat, $u = (1,1,1,1,1)$. Differences $1,0,0,0,0,-1$: numerator $2$, denominator $5$. $R = \tfrac25\tfrac km$, so $\omega_1 \le 0.632456\sqrt{k/m}$.

Trial 2 · tent, $u = (1,2,3,2,1)$. Differences $1,1,1,-1,-1,-1$: numerator $6$, denominator $19$. $R = \tfrac{6}{19}\tfrac km$, so $$\omega_1 \le \sqrt{\tfrac{6}{19}}\sqrt{k/m} = 0.561951\sqrt{k/m}.$$

Trial 3 · rounded tent, $u = (2,4,5,4,2)$. Differences $2,2,1,-1,-2,-2$: numerator $18$, denominator $65$. $R = \tfrac{18}{65}\tfrac km$, so $\omega_1 \le 0.526235\sqrt{k/m}$.

The exact answer. $\omega_1 = 2\sin(\pi/12)\sqrt{k/m} = \sqrt{2-\sqrt3}\,\sqrt{k/m} = 0.517638\sqrt{k/m}$.

trial	$\omega_1$ bound	excess
flat	0.632456	22.18%
tent	0.561951	8.56%
rounded	0.526235	1.66%
exact	0.517638	—

The flat trial violates the boundary behaviour badly (a jump of $1$ at each wall) and still lands within $22\%$. That is the practical point: Rayleigh's method is hard to make fail.

All numbers checked. Exact: $\lambda_1 = 2-\sqrt3 = 0.2679492$, $\omega_1 = 0.5176381$. Flat: $R = 0.4$, $\sqrt R = 0.6324555$, excess $22.181\%$. Tent: $R = 6/19 = 0.3157895$, $\sqrt R = 0.5619515$, excess $8.5607\%$. Rounded: $R = 18/65 = 0.2769231$, $\sqrt R = 0.5262348$, excess $1.6608\%$. The full spectrum of $A_5$ is $2\sin(n\pi/12)$ for $n = 1,\dots,5$, i.e. $0.5176381,\,1,\,1.4142136,\,1.7320508,\,1.9318517$, confirmed by numerical diagonalisation.

Computing the numerator without touching $K$. Note that the sum-of-squares form $u^{\!\top}A_Nu = \sum_{j=0}^{N}(u_{j+1}-u_j)^2$ was used throughout, and no matrix multiplication was performed anywhere on this slide. That is typical: the quadratic form is usually the energy, and the energy of a guessed shape is something you can write down from physics. Rayleigh's own applications, in The Theory of Sound (1877), were to strings, bars, plates and air columns, long before matrices were a standard tool.

Bounding from below is much harder. Nothing here gives a lower bound on $\omega_1$, and none of the elementary arguments does. The standard device is Temple's inequality, which needs a lower estimate of $\lambda_2$ as well as the trial vector; it is stated and proved in the solution to P13. (Weinstein's method is another, named here and not used.) The trace looks as if it ought to help, $\sum_p\lambda_p = \mathrm{tr}(M^{-1}K) = 2Nk/m$ for the fixed chain, but a smallest element is at most the average, so it gives $\lambda_1 \le 2k/m$ — another upper bound, which is the recurring difficulty.

Where this is used for real. Every structural engineer's first estimate of a building's fundamental sway frequency is a Rayleigh quotient with an assumed mode shape, and a conforming finite-element model with a consistent mass matrix is minimising $R$ over the subspace its mesh can represent — which is why its frequencies come out above the true ones (the min–max theorem, applied to a subspace), and why refining the mesh, which enlarges the subspace, can only lower them.

07 · Rayleigh · example

The trial shapes, drawn

every bound is above, as promised

The true mode is $v_j = \sin(j\pi/6) \propto (1,\,1.732,\,2,\,1.732,\,1)$. The tent $(1,2,3,2,1)$ is too peaked; $(2,4,5,4,2)$ is closer, and its error in $\omega_1$ is $1.66\%$ while its shape error is $7.2\%$ — the quadratic insensitivity, visible.

"Shape error" here is the distance from the true mode after the best rescaling, relative to its length: $27.7\%$ for flat, $15.2\%$ for the tent, $7.2\%$ for the rounded tent. The frequency errors are $22.2\%$, $8.6\%$ and $1.7\%$ — shrinking much faster than the shape errors.

The exact lowest mode of five masses (solid, a sampled sine) and the three trials, each rescaled to fit it as well as it can. The rounded tent is nearly on top of it, and its bound is $1.7\%$ high.

How the trials were rescaled. Rayleigh's quotient ignores scale, so to compare shapes each trial $t$ is multiplied by the $c$ minimising $\lVert ct - v\rVert$, namely $c = t\cdot v/t\cdot t$, with $v_j = \sin(j\pi/6)$. The shape error is then $\lVert ct - v\rVert/\lVert v\rVert$: $0.2774$, $0.1520$, $0.0720$. Squared, these are $0.077$, $0.023$ and $0.0052$, and the excesses in $\lambda_1 = \omega_1^2$ are $49.3\%$, $17.9\%$ and $3.35\%$ — roughly proportional to the squared shape error, the constant of proportionality being set by how far the higher $\lambda_p$ sit above $\lambda_1$.

What to look for. The flat trial is the worst at the walls, where it jumps from $0$ to its full height in one spacing; that jump stretches the end springs hard and is most of its excess energy. The tent is too pointed in the middle. The rounded tent's only visible fault is a slightly low shoulder.

07 · closing

The chain, closed — and what it unlocks

The chain, from start to finish:

$M = mI$, $K = kA_N$ from the energy alone;
$K$ positive definite by a sum of squares, so every $\omega_n > 0$ before any computation;
$v^{(n)}_j = \sin\big(jn\pi/(N{+}1)\big)$, $\omega_n = 2\sqrt{k/m}\sin\big(n\pi/2(N{+}1)\big)$, exactly, with the ansatz verified rather than asserted;
the walls removed: one frequency drops to zero, and the symmetry responsible is named;
its fundamental bounded to $1.7\%$ by a vector written down in five seconds.

And a number worth keeping: the continuum theory of unit 18 will predict this chain's fundamental $0.093\%$ too high, which this unit knew in advance.

What this unlocks.

18: $a \to 0$, and $A_N/a^2$ becomes $-\partial_x^2$: the wave equation, with this unit's $\omega(\kappa)$ as its error bar.
19: the modes are a basis — here by the spectral theorem in $\mathbb R^n$, there by Sturm–Liouville theory in $L^2$, and the $M$-inner product becomes a weighted integral.
16: the inertia tensor's principal axes are this unit's theorem again, with $K$ replaced by a moment-of-inertia form.
13: symmetry $\Rightarrow$ zero mode becomes symmetry $\Rightarrow$ conserved quantity.
22: principal stresses; 32: diagonalising a metric at a point.

The one sentence to keep: a pencil $(K, M)$ of symmetric forms with $M$ positive definite behaves exactly like a single symmetric matrix, because $M^{1/2}$ turns it into one. Everything else in this unit is a consequence, and every later appearance of "principal axes", "normal modes", "eigenfunctions" or "diagonalise the metric at a point" is that sentence with different nouns.

If one computation survives, make it the sine ansatz of “Worked: a sine is an eigenvector”. It is three lines, it is exact for every $N$, and it is the discrete original of every standing-wave argument in units 18, 19, 23 and 24. Lagrange had it in 1759, nearly a century before matrices were named, and he used it as this unit does: letting $N$ grow to reach the continuous string, the limit unit 18 takes properly.

07 · checkpoint

Done when

checkpoint

Given any pair $(M, K)$ of symmetric matrices with $M$ positive definite, you can:

produce the normal frequencies and mode shapes, by the reduction to $S = M^{-1/2}KM^{-1/2}$ or directly from $\det(K-\lambda M) = 0$;
verify orthogonality in the $M$-inner product, and say why the Euclidean one is the wrong test;
identify every zero mode and name the symmetry of $V$ responsible for it;
explain why $\omega^2$ is always real, and why an $\omega$ with nonzero real and imaginary parts cannot occur.

Falsifiable four ways: checking orthogonality with the Euclidean product, missing a zero mode, claiming a complex frequency, or failing to say where positive definiteness was used. Record it with ./mr done 07.

Problems

./mr problems 07 — 13 problems, about five hours. Hints and worked solutions fold under each problem on the same page.

08 the central-force problem (needs 04, 06) · 12 Lagrangian mechanics, where a general $M(q)$ arrives properly · 18 the continuum limit of this unit's chain · 19 the modes as a complete basis.

Try the checkpoint cold, on paper, before looking at anything. A good self-test: three masses $m, 2m, m$ in a line joined by two springs $k$, with no walls. You should get one zero mode $(1,1,1)$ from translation invariance, and two others; the answers are $\omega^2 = 0,\ k/m,\ 2k/m$, with shapes $(1,1,1)$, $(1,0,-1)$ and $(1,-1,1)$. Check that the three are $M$-orthogonal with $M = \mathrm{diag}(m,2m,m)$ — note that $(1,1,1)$ and $(1,-1,1)$ are not Euclidean-orthogonal, their dot product being $1$, while their $M$-product is $m - 2m + m = 0$.

If the derivation stalls, the usual place is the reduction: remember that $M^{-1}K$ is the wrong object and $M^{-1/2}KM^{-1/2}$ is the right one, and that the eigenvectors transform by $v = M^{-1/2}w$, not by $w$ itself. “$M^{1/2}$, and the problem made symmetric” is the one to reread.

Unit 18 picks this up directly: it takes $N \to \infty$ with $Na$ fixed and turns the chain into a string, at which point the sum over modes becomes a Fourier series and the second-difference matrix becomes $-\partial_x^2$. Everything proved here about orthogonality and completeness has a counterpart there, and the counterpart is harder only because the space is infinite-dimensional.

Many oscillators, and the spectral theorem earning its keep

One chain, five sections

$n$ coordinates, one linear system

Two quadratic forms fall out of the physics

A mass matrix that is not diagonal

First order, in $2n$ dimensions

The matrix exponential

When $A$ is diagonalisable

The degenerate case: Jordan, and $t\,e^{\lambda t}$

Critical damping, plotted

Two quadratic forms, one basis

The generalised eigenvalue problem

Three tempting routes, all wrong

The spectral theorem, and its variational proof

Two things the proof leaves behind

$M^{1/2}$, and the problem made symmetric

What the reduction proves, all at once

The inner product the modes are orthogonal in

Worked: two masses, and the orthogonality that is not Euclidean

Two masses, two geometries

Normal coordinates: the equations fall apart

One release, two clean oscillations

The chain, exactly

Writing down $M$ and $K$

Worked: a sine is an eigenvector

Four checks on the formula

The dispersion curve, and where it bends

A continuum theory's own error bar

The deficit, mode by mode

Zero modes, and the symmetries responsible

Cut the chain loose

The free chain's three modes

A symmetry of $V$ is a null vector of $K$

Counting zero modes

What a zero mode actually does

The three regimes, side by side

Rayleigh: bounding a frequency without solving

The Rayleigh quotient

The quotient for two masses, drawn

Worked: bounding the fundamental of a five-mass chain

The trial shapes, drawn

The chain, closed — and what it unlocks

Done when

Problems

Next